Rd Sharma 2022 for Class 9 Maths Chapter 3 - Rationalisation

Answer:

(i) We know that $\sqrt[n]{a}\times \sqrt[n]{b}=\sqrt[n]{ab}$. We will use this property to simplify the expression $\sqrt[3]{4}\times \sqrt[3]{16}$.

$\therefore \sqrt[3]{4}\times \sqrt[3]{16}$

Hence the value of the given expression is 4.

(ii) We know that $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$. We will use this property to simplify the expression $\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}$.

Hence the value of the given expression is 5.

(iii)
$\begin{array}{rcl}4\sqrt{28}÷3\sqrt{7}& =& \frac{4\sqrt{28}}{3\sqrt{7}}\\ & =& \frac{4\sqrt{4\times 7}}{3\sqrt{7}}\\ & =& \frac{4\times 2\times \sqrt{7}}{3\sqrt{7}}\\ & =& \frac{8}{3}\end{array}$

Hence, the value of the expression is $\frac{8}{3}$.

Answer:

(i) We can simplify the expression as 

Hence the value of the expression is

(ii) We can simplify the expression as 

Hence the value of the expression is

(iii) We can simplify the expression as 

Hence the value of the expression is .

Answer:

We know that $\left(a+b\right)\left(a-b\right)=a^2-b^2$.
(i)

Hence the value of the expression is 110.

(ii)

Hence the value of the expression is 18.

(iii)

Hence the value of the expression is 6

(iv)

Hence the value of the expression is 6.

(v)

Hence the value of the expression is 3.

(vi)
$$\begin{gathered} \therefore \left(3\sqrt{5}-5\sqrt{2}\right)\left(4\sqrt{5}+3\sqrt{2}\right) \\ =3\sqrt{5}\times 4\sqrt{5}+3\sqrt{5}\times 3\sqrt{2}-5\sqrt{2}\times 4\sqrt{5}-5\sqrt{2}\times 3\sqrt{2} \\ =60+9\sqrt{10}-20\sqrt{10}-30 \\ =30-11\sqrt{10} \end{gathered}$$

Answer:

(i) We know that. We will use this property to simplify the expression.

Hence the value of expression is

(ii) We know that. We will use this property to simplify the expression.

Hence the value of expression is

(iii) We know that. We will use this property to simplify the expression.

Hence the value of expression is.

Answer:

(i) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(ii) We know that rationalization factor foris. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(iii) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(iv) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(v) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(vi) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(vii) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

Answer:

(i) We know that rationalization factor of the denominator is. We will multiply numerator and denominator of the given expression by, to get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(ii) We know that rationalization factor of the denominator is . We will multiply numerator and denominator of the given expression by , to get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(iii) We know that rationalization factor of the denominator is. We will multiply numerator and denominator of the given expression by, to get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(iv) We know that rationalization factor of the denominator is. We will multiply numerator and denominator of the given expression by, to get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(v) Given that

Putting the value of, we get 

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(vi) We know that rationalization factor of the denominator is. We will multiply numerator and denominator of the given expression by, to get

Putting the value of and, we get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

Answer:

(i) Multiply numerator and denominator by $\sqrt{2}-\sqrt{3}$
$$\begin{gathered} =\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{{\left(\sqrt{2}\right)}^2-{\left(\sqrt{3}\right)}^2} \\ =\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{-1}=\sqrt{18}-\sqrt{12} \\ =3\sqrt{2}-2\sqrt{3} \end{gathered}$$

(ii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with a rational denominator to.

(iii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(iv) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(v) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(vi) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(vii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(viii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(ix) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to .

(x) Multiply numerator denominator by $2+\sqrt{3}$
$\begin{array}{rcl}\therefore \frac{2+\sqrt{3}}{2-\sqrt{3}}& =& \frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}\\ & =& \frac{{\left(2+\sqrt{3}\right)}^2}{2^2-{\left(\sqrt{3}\right)}^2} [\mathrm{Using} a^2-b^2=(a-b) (a+b\left)\right]\\ & =& \frac{2^2+{\left(\sqrt{3}\right)}^2+2\times 2\sqrt{3}}{4-3}\\ & =& 7+4\sqrt{3} [\mathrm{Using} {(a+b)}^2=a^2+b^2+2ab]\end{array}$

Answer:

(i) We know that rationalization factor for is . We will multiply the numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(ii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(iii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(iv) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(v) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(vi) We know that rationalization factor for $\sqrt{5}-\sqrt{3}$ is $\sqrt{5}+\sqrt{3}$ . We will multiply numerator and denominator of the given expression $\begin{array}{rcl}& & \frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\end{array}$ by$\sqrt{5}+\sqrt{3}$, to get

$\begin{array}{rcl}& \Rightarrow & \frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\\ & =& \frac{\left(3\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}\\ & =& \frac{15+3\sqrt{15}+\sqrt{15}+3}{5-3}\\ & =& \frac{18+4\sqrt{15}}{2}\\ & =& 9+2\sqrt{15}\end{array}$

Hence the given expression is simplified to $9+2\sqrt{15}$.

Answer:

(i) We know that rationalization factor forand areand respectively. We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to.

(ii) We know that rationalization factor forand areand respectively. We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to.

(iii) We know that rationalization factor forand areand respectively. We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to.

Answer:

(i) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get 

Hence, we get.

(ii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get 

Hence, we get.

(iii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get 

Hence, we get. 

(iv) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get 

Hence, we get.

(v) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get 

Hence, we get.

(vi) Given, 
Rationalizing the denominator on LHS by multiplying the numerator and denominator by 


On equating rational and irrational terms, we get 

Answer:

We know that rationalization factor for is . We will multiply denominator and numerator of the given expression by , to get

Putting the values of and, we get

Hence value of the given expression is.

Answer:

(i) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Putting the values of, we get 

Hence the given expression is simplified to.

(ii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Putting the value of, we get 

Hence the given expression is simplified to.

(iii)
$$\begin{gathered} \frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}} \\ =\frac{4\left(3\sqrt{3}+2\sqrt{2}\right)+3\left(3\sqrt{3}-2\sqrt{2}\right)}{\left(3\sqrt{3}-2\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)} \\ =\frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{{\left(3\sqrt{3}\right)}^2-{\left(2\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{21\sqrt{3}+2\sqrt{2}}{27-8} \\ =\frac{21\sqrt{3}+2\sqrt{2}}{19} \\ =\frac{21\times 1.732+2\times 1.414}{19} \\ =\frac{36.372+2.828}{19} \\ =\frac{39.2}{19} \\ =2.063 \end{gathered}$$
 

Answer:

(i) We know that rationalization factor forand areand respectively.
We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to 11.

(ii) We know that rationalization factor for and areand respectively.
We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to $\sqrt{5}$.

(iii) Given,
$\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$
Rationalizing the denominators of the terms
$\left[\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}\times \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}\right]-\left[\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}\right]-\left[\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}\times \frac{\sqrt{15}-3\sqrt{2}}{\sqrt{15}-3\sqrt{2}}\right]$
[Using identity (a – b)(a + b) = a² – b²]
$$\begin{gathered} =\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{{\left(\sqrt{10}\right)}^2-{\left(\sqrt{3}\right)}^2}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{5}\right)}^2}-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{{\left(\sqrt{15}\right)}^2-{\left(3\sqrt{2}\right)}^2} \\ =\frac{7\left(\sqrt{30}-3\right)}{10-3}-\frac{\left(2\sqrt{30}-10\right)}{6-5}-\frac{\left(3\sqrt{30}-18\right)}{15-18} \\ =\left(\sqrt{30}-3\right)-\left(2\sqrt{30}-10\right)+\left(\sqrt{30}-6\right) \\ =\sqrt{30}-3-2\sqrt{30}+10+\sqrt{30}-6 \\ =1 \end{gathered}$$

Answer:

We know that. We have to find the value of.

As therefore, 

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Putting the value of and , we get

Hence the value of the given expression

Answer:

We know that. We have to find the value of . As therefore,

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Putting the value of x and , we get

Hence the given expression is simplified to.

Answer:

We have,

It can be simplified as 

On squaring both sides, we get 

The given equation can be rewritten as.

Therefore, we have 

Hence, the value of given expression is.

Answer:

Given, $a=\frac{3+\sqrt{5}}{2}$
then $\frac{1}{a}=\frac{2}{3+\sqrt{5}}$
Rationalizing the denominator, we have
$\begin{array}{rcl}\frac{1}{a}& =& \frac{2}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}\\ & =& \frac{6-2\sqrt{5}}{3^2-{\left(\sqrt{5}\right)}^2}\\ & =& \frac{6-2\sqrt{5}}{9-5}\\ & =& \frac{6-2\sqrt{5}}{4}\\ & =& \frac{3-\sqrt{5}}{2}\end{array}$

$\begin{array}{rcl}\therefore a^2+\frac{1}{a^2}& =& a^2+\frac{1}{a^2}+2-2 [\mathrm{add} \mathrm{and} \mathrm{subtract} 2]\\ & =& {\left(a+\frac{1}{a}\right)}^2-2\\ & =& {\left(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}\right)}^2-2\\ & =& {\left(\frac{6}{2}\right)}^2-2\\ & =& 3^2-2\\ & =& 9-2\\ & =& 7\end{array}$

Hence, the value of $a^2+\frac{1}{a^2}$ is 7.

Answer:

Given: $a=5+2\sqrt{6} \mathrm{and} b=\frac{1}{a}$

$\begin{array}{rcl}\therefore b& =& \frac{1}{5+2\sqrt{6}}\\ & =& \frac{1}{5+2\sqrt{6}}\times \frac{5-2\sqrt{6}}{5-2\sqrt{6}}\\ & =& \frac{5-2\sqrt{6}}{{\left(5\right)}^2-{\left(2\sqrt{6}\right)}^2}\\ & =& \frac{5-2\sqrt{6}}{25-24}\\ & =& 5-2\sqrt{6}\end{array}$

$\begin{array}{rcl}\therefore a^2+b^2& =& {\left(a+b\right)}^2-2ab\\ & =& {\left(5+2\sqrt{6}+5-2\sqrt{6}\right)}^2-2\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\\ & =& {\left(10\right)}^2-2\left[5^2-{\left(2\sqrt{6}\right)}^2\right]\\ & =& 100-2\left(25-24\right)\\ & =& 98\end{array}$

Hence, the value of a² + b² is 98.

Answer:

$\begin{array}{rcl}\frac{6}{3\sqrt{2}-2\sqrt{3}}& =& \frac{6}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}\\ & =& \frac{6\left(3\sqrt{2}+2\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}\\ & =& \frac{6\left(3\sqrt{2}+2\sqrt{3}\right)}{{\left(3\sqrt{2}\right)}^2-{\left(2\sqrt{3}\right)}^2}\\ & =& \frac{6\left(3\sqrt{2}+2\sqrt{3}\right)}{18-12}\\ & =& 3\sqrt{2}+2\sqrt{3}\\ & =& 3\sqrt{2}-\left(-2\right)\sqrt{3}\end{array}$

Thus, the value of a is −2.

Answer:

Given that 

It can be simplified as 

Hence the value of the given expression is.

Answer:

Given that, it’s reciprocal is given as 

It can be simplified by rationalizing the denominator. The rationalizing factor of is, we will multiply numerator and denominator of the given expression by, to get

Hence reciprocal of the given expression is.

Answer:

The rationalizing factor of is. Hence the rationalizing factor of is .

Answer:

It is given that;

.we need to find x and y

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get 

Hence, we get.

Answer:

Given that.Hence is given as

We know that rationalization factor for is . We will multiply each side of the given expression by, to get

Hence the value of the given expression is.

Answer:

Given that, hence is given as

.we are asked to find

We know that rationalization factor for is . We will multiply each side of the given expression by, to get

Therefore,

Hence value of the given expression is.

Answer:

Given that, hence $\frac{1}{x}$is given as

.We are asked to find

We know that rationalization factor for is . We will multiply each side of the given expression by, to get

Therefore,

Hence value of the given expression is.

Answer:

Given that, we know that rationalization factor of is

So the rationalization factor of is.

Answer:

We are asked to simplify. It can be written in the form as

Hence the value of given expression is.

Answer:

We are asked to simplify. It can be written in the form as

Hence the value of the given expression is.

Answer:

Given that:.It can be written in the form as

Therefore,

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence,

Therefore, value of the given expression is.

Answer:

$$\begin{gathered} \frac{1}{\sqrt{7}+2} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{by} \left(\sqrt{7}-2\right), \mathrm{we} \mathrm{get} \\ =\frac{1}{\sqrt{7}+2}\times \frac{\sqrt{7}-2}{\sqrt{7}-2} \\ =\frac{\sqrt{7}-2}{{\left(\sqrt{7}\right)}^2-2^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ =\frac{\sqrt{7}-2}{7-4} \\ =\frac{\sqrt{7}-2}{3} \end{gathered}$$

Hence, the number obtained by rationalizing the denominator of $\frac{1}{\sqrt{7}+2}$ is $\underline{ \frac{\sqrt{7}-2}{3} }.$.

Answer:

$$\begin{gathered} \frac{1}{\sqrt{9}-\sqrt{8}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{by} \left(\sqrt{9}+\sqrt{8}\right), \mathrm{we} \mathrm{get} \\ =\frac{1}{\sqrt{9}-\sqrt{8}}\times \frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}} \\ =\frac{\sqrt{9}+\sqrt{8}}{{\left(\sqrt{9}\right)}^2-{\left(\sqrt{8}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ =\frac{\sqrt{9}+\sqrt{8}}{9-8} \\ =\frac{\sqrt{9}+\sqrt{8}}{1} \\ =\sqrt{9}+\sqrt{8} \\ =3+2\sqrt{2} \\ \mathrm{Now}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \\ \frac{1}{\sqrt{9}-\sqrt{8}}=A+B\sqrt{2} \\ \Rightarrow 3+2\sqrt{2}=A+B\sqrt{2} \\ \Rightarrow A=3 \mathrm{and} B=2 \end{gathered}$$


Hence, if $\frac{1}{\sqrt{9}-\sqrt{8}}=A+B\sqrt{2}$, then A = 3 and B = 2.

Answer:

$$\begin{gathered} \frac{7}{3\sqrt{3}-2\sqrt{2}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{by} \left(3\sqrt{3}+2\sqrt{2}\right), \mathrm{we} \mathrm{get} \\ =\frac{7}{3\sqrt{3}-2\sqrt{2}}\times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}} \\ =\frac{7\left(3\sqrt{3}+2\sqrt{2}\right)}{{\left(3\sqrt{3}\right)}^2-{\left(2\sqrt{2}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ =\frac{21\sqrt{3}+14\sqrt{2}}{27-8} \\ =\frac{21\sqrt{3}+14\sqrt{2}}{19} \end{gathered}$$


Hence, after rationalizing the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$, we get the denominator as $\underline{\frac{21\sqrt{3}+14\sqrt{2}}{19}}.$

Answer:

$$\begin{gathered} \mathrm{Given}: a=2+\sqrt{3} \\ \mathrm{Now}, \frac{1}{a}=\frac{1}{2+\sqrt{3}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(2-\sqrt{3}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{1}{a}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ \Rightarrow \frac{1}{a}=\frac{2-\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{a}=\frac{2-\sqrt{3}}{4-3} \\ \Rightarrow \frac{1}{a}=\frac{2-\sqrt{3}}{1} \\ \Rightarrow \frac{1}{a}=2-\sqrt{3} \\ \mathrm{Thus}, \\ a-\frac{1}{a}=\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right) \\ =2+\sqrt{3}-2+\sqrt{3} \\ =2\sqrt{3} \end{gathered}$$


Hence, if $a=2+\sqrt{3}, \mathrm{then} a-\frac{1}{a}=\underline{2\sqrt{3}}.$

Answer:

$$\begin{gathered} \mathrm{Given}: a=5+2\sqrt{6} \\ \mathrm{Now}, \frac{1}{a}=\frac{1}{5+2\sqrt{6}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(5-2\sqrt{6}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{1}{a}=\frac{1}{5+2\sqrt{6}}\times \frac{5-2\sqrt{6}}{5-2\sqrt{6}} \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{{\left(5\right)}^2-{\left(2\sqrt{6}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{25-24} \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{1} \\ \Rightarrow \frac{1}{a}=5-2\sqrt{6} \\ \mathrm{Thus}, \\ a+\frac{1}{a}=\left(5+2\sqrt{6}\right)+\left(5-2\sqrt{6}\right) \\ =5+2\sqrt{6}+5-2\sqrt{6} \\ =10 \end{gathered}$$


Hence, if $a=5+2\sqrt{6}, \mathrm{then} a+\frac{1}{a}=\underline{10}.$

Answer:

$$\begin{gathered} \mathrm{Given}: x=\sqrt{6}+\sqrt{5} \\ \mathrm{Now}, \frac{1}{x}=\frac{1}{\sqrt{6}+\sqrt{5}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{6}-\sqrt{5}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{1}{x}=\frac{1}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{5}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{1} \\ \Rightarrow \frac{1}{x}=\sqrt{6}-\sqrt{5} \\ \mathrm{Thus}, \\ {x}^2+\frac{1}{x^2}-2={\left(x-\frac{1}{x}\right)}^2 \\ ={\left(\left(\sqrt{6}+\sqrt{5}\right)-\left(\sqrt{6}-\sqrt{5}\right)\right)}^2 \\ ={\left(\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}\right)}^2 \\ ={\left(2\sqrt{5}\right)}^2 \\ =20 \end{gathered}$$


Hence, if $x=\sqrt{6}+\sqrt{5}, \mathrm{then} x^2+\frac{1}{x^2}-2=\underline{20}.$

Answer:

$$\begin{gathered} \mathrm{Given}: x=3-\sqrt{8} \\ \mathrm{Now}, \frac{1}{x}=\frac{1}{3-\sqrt{8}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(3+\sqrt{8}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow \frac{1}{x}=\frac{1}{3-\sqrt{8}}\times \frac{3+\sqrt{8}}{3+\sqrt{8}} \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{{\left(3\right)}^2-{\left(\sqrt{8}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{9-8} \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{1} \\ \Rightarrow \frac{1}{x}=3+\sqrt{8} \\ \mathrm{Thus}, \\ {\left(x-\frac{1}{x}\right)}^2={\left(\left(3-\sqrt{8}\right)-\left(3+\sqrt{8}\right)\right)}^2 \\ ={\left(3-\sqrt{8}-3-\sqrt{8}\right)}^2 \\ ={\left(-2\sqrt{8}\right)}^2 \\ =32 \end{gathered}$$


Hence, if $x=3-\sqrt{8}, \mathrm{then} {\left(x-\frac{1}{x}\right)}^2=\underline{32}.$

Answer:

$$\begin{gathered} \mathrm{Given}: \\ x=\frac{2}{\sqrt{3}-\sqrt{5}} \\ y=\frac{2}{\sqrt{3}+\sqrt{5}} \\ \mathrm{Now}, \\ x=\frac{2}{\sqrt{3}-\sqrt{5}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{3}+\sqrt{5}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow x=\frac{2}{\sqrt{3}-\sqrt{5}}\times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}} \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{5}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{3-5} \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{-2} \\ \Rightarrow x=\frac{-\left(\sqrt{3}+\sqrt{5}\right)}{1} \\ \Rightarrow x=-\sqrt{3}-\sqrt{5} \\ y=\frac{2}{\sqrt{3}+\sqrt{5}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{3}-\sqrt{5}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow y=\frac{2}{\sqrt{3}+\sqrt{5}}\times \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{5}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{3-5} \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{-2} \\ \Rightarrow y=\frac{-\left(\sqrt{3}-\sqrt{5}\right)}{1} \\ \Rightarrow y=-\sqrt{3}+\sqrt{5} \\ \mathrm{Thus}, \\ x+y=\left(-\sqrt{3}-\sqrt{5}\right)+\left(-\sqrt{3}+\sqrt{5}\right) \\ =\left(-\sqrt{3}-\sqrt{5}-\sqrt{3}+\sqrt{5}\right) \\ =\left(-2\sqrt{3}\right) \\ =-2\sqrt{3} \end{gathered}$$


Hence, x + y = $\underline{-2\sqrt{3}}$.

Answer:

$$\begin{gathered} \mathrm{Given}: \\ \sqrt{5+2\sqrt{6}}=A+\sqrt{3} \\ \mathrm{Now}, \\ \sqrt{5+2\sqrt{6}}=A+\sqrt{3} \\ \Rightarrow \sqrt{2+3+2\sqrt{6}}=A+\sqrt{3} \\ \Rightarrow \sqrt{{\left(\sqrt{2}\right)}^2+{\left(\sqrt{3}\right)}^2+2\sqrt{6}}=A+\sqrt{3} \\ \Rightarrow \sqrt{{\left(\sqrt{2}+\sqrt{3}\right)}^2}=A+\sqrt{3} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}{\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow \sqrt{2}+\sqrt{3}=A+\sqrt{3} \\ \Rightarrow A=\sqrt{2} \end{gathered}$$


Hence, if $\sqrt{5+2\sqrt{6}}=A+\sqrt{3}, \mathrm{then} A=\underline{\sqrt{2}}.$

Answer:

$$\begin{gathered} \sqrt{7+2\sqrt{6}} \\ =\sqrt{6+1+2\sqrt{6}} \\ =\sqrt{{\left(\sqrt{6}\right)}^2+{\left(1\right)}^2+2\sqrt{6}} \\ =\sqrt{{\left(\sqrt{6}+1\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}{\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ =\sqrt{6}+1 ....\left(1\right) \\ \sqrt{7-2\sqrt{6}} \\ =\sqrt{6+1-2\sqrt{6}} \\ =\sqrt{{\left(\sqrt{6}\right)}^2+{\left(1\right)}^2-2\sqrt{6}} \\ =\sqrt{{\left(\sqrt{6}-1\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}{\left(a-b\right)}^2=a^2+b^2-2ab\right) \\ =\sqrt{6}-1 ....\left(2\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right) \\ \sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}=\left(\sqrt{6}+1\right)-\left(\sqrt{6}-1\right) \\ =\sqrt{6}+1-\sqrt{6}+1 \\ =2 \end{gathered}$$


Hence, $\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}=\underline{2}.$

Answer:

$$\begin{gathered} \mathrm{Given}: \\ x=\frac{2}{\sqrt{10}-\sqrt{8}} \\ y=\frac{2}{\sqrt{10}+2\sqrt{2}} \\ \mathrm{Now}, \\ x=\frac{2}{\sqrt{10}-\sqrt{8}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{10}+\sqrt{8}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow x=\frac{2}{\sqrt{10}-\sqrt{8}}\times \frac{\sqrt{10}+\sqrt{8}}{\sqrt{10}+\sqrt{8}} \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{{\left(\sqrt{10}\right)}^2-{\left(\sqrt{8}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{10-8} \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{2} \\ \Rightarrow x=\frac{\left(\sqrt{10}+\sqrt{8}\right)}{1} \\ \Rightarrow x=\sqrt{10}+\sqrt{8} \\ \Rightarrow x=\sqrt{10}+2\sqrt{2} \\ y=\frac{2}{\sqrt{10}+2\sqrt{2}} \\ \mathrm{Multiply} \mathrm{and} \mathrm{divide} \mathrm{RHS} \mathrm{by} \left(\sqrt{10}-2\sqrt{2}\right), \mathrm{we} \mathrm{get} \\ \Rightarrow y=\frac{2}{\sqrt{10}+2\sqrt{2}}\times \frac{\sqrt{10}-2\sqrt{2}}{\sqrt{10}-2\sqrt{2}} \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{{\left(\sqrt{10}\right)}^2-{\left(2\sqrt{2}\right)}^2} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{10-8} \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{2} \\ \Rightarrow y=\frac{\left(\sqrt{10}-2\sqrt{2}\right)}{1} \\ \Rightarrow y=\sqrt{10}-2\sqrt{2} \\ \mathrm{Thus}, \\ {\left(x-y\right)}^2={\left(\left(\sqrt{10}+2\sqrt{2}\right)-\left(\sqrt{10}-2\sqrt{2}\right)\right)}^2 \\ ={\left(\sqrt{10}+2\sqrt{2}-\sqrt{10}+2\sqrt{2}\right)}^2 \\ ={\left(4\sqrt{2}\right)}^2 \\ =32 \end{gathered}$$


Hence, if $x=\frac{2}{\sqrt{10}-\sqrt{8}} \mathrm{and} y=\frac{2}{\sqrt{10}+2\sqrt{2}}, \mathrm{then} {(x-y)}^2=\underline{32}.$