Solution

We know that every real number is either an irrational number or rational number. Therefore, every rational number is a real number.

Hence, the correct option is (C).

Solution

Between two rational numbers there are infinitely many rational numbers.

Hence, the correct option is (C).

Solution

We know that, the decimal representation of a rational number cannot be non-terminating and non-repeating.

Hence, the correct option is (D).

Solution

We know that, the product of any two irrational numbers is sometimes rational and sometimes irrational.

Hence, the correct option is (D).

Solution

The decimal expansion of the number $\sqrt{2}$ is $1.41421 \ldots$, which is non-terminating and nonrecurring.

Hence, the correct option is (B).

Solution

(A) $\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}$, Which is rational number.

(B) $\dfrac{\sqrt{12}}{\sqrt{3}}=\dfrac{\sqrt{4 \times 3}}{\sqrt{3}}=\dfrac{2 \sqrt{3}}{\sqrt{3}}=2$, Which is rational number.

(C) $\sqrt{7}$ is a irrational number.

(D) $\sqrt{81}=\sqrt{9^{2}}=9$, which is a rational number.

Hence, the correct option is (C).

Solution

(A) 0.14 is a terminating decimal. Hence, it can’t be an irrational number.

(B) $0.14 \overline{16}$ is a non-terminating and recurring decimal. Hence, it can’t be an irrational number.

(C) $0 . \overline{1416}$ is a non-terminating and recurring decimal. Hence, it can’t be an irrational number.

(D) $0.4014001400014 \ldots$ is a non-terminating and non-recurring decimal. Hence, it is an irrational number.

Hence, the correct option is (D).

Solution

We know that,

$\sqrt{2}=1.4142135$ and $\sqrt{3}=1.732050807$

5 is a rational number which lies between $\sqrt{2}=1.4142135$ and $\sqrt{3}=1.732050807$.

Hence, the correct option is (C).

Solution

Let $x=1.999 \ldots=1 . \overline{9} \quad \ldots$ (I)

Then, $10 x=19.999 \ldots=19 . \overline{9}$

Subtracting (I) and (II), get:

$9 x=18$

$x=2$

Therefore, the value of $1.999 \ldots$ in the form $\dfrac{p}{q}$ is 2 or $\dfrac{2}{1}$.

Hence, the correct option is (C).

Solution

$2 \sqrt{3}+\sqrt{3}=3 \sqrt{3}$

Hence, the correct option is (C).

Solution

$\sqrt{10} \times \sqrt{15}=\sqrt{5 \times 2 \times 5 \times 3}=5 \sqrt{6}$

Hence, the correct option is (B).

Solution

Rationalizing the denominator as follows:

$ \begin{aligned} \dfrac{1}{\sqrt{7}-2} & =\dfrac{1}{\sqrt{7}-2} \times \dfrac{\sqrt{7}+2}{\sqrt{7}+2} \\ & =\dfrac{\sqrt{7}+2}{(\sqrt{7})^{2}-2^{2}} \\ & =\dfrac{\sqrt{7}+2}{7-4} \\ & =\dfrac{\sqrt{7}+2}{3} \end{aligned} $

Hence, the correct option is (A).

Solution

$ \begin{aligned} \dfrac{1}{\sqrt{9}-\sqrt{8}} & =\dfrac{1}{\sqrt{9}-\sqrt{8}} \times \dfrac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}} \\ & =\dfrac{\sqrt{9}+\sqrt{8}}{(\sqrt{9})^{2}-(\sqrt{8})^{2}} \\ & =\dfrac{\sqrt{3^{2}}+\sqrt{2^{3}}}{9-8} \\ & =3+2 \sqrt{2} \end{aligned} $

Hence, the correct option is (D).

Solution

$ \begin{aligned} \dfrac{7}{3 \sqrt{3}-2 \sqrt{2}} & =\dfrac{7}{3 \sqrt{3}-2 \sqrt{2}} \times \dfrac{3 \sqrt{3}+2 \sqrt{2}}{3 \sqrt{3}+2 \sqrt{2}} \\ & =\dfrac{7(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}} \\ & =\dfrac{7(3 \sqrt{3}+2 \sqrt{2})}{27-8} \\ & =\dfrac{7(3 \sqrt{3}+2 \sqrt{2})}{19} \end{aligned} $

Hence, the correct option is (B).

Solution

$ \begin{aligned} \dfrac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}} & =\dfrac{\sqrt{16 \times 2}+\sqrt{16 \times 3}}{\sqrt{4 \times 2}+\sqrt{4 \times 3}} \\ & =\dfrac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}} \\ & =\dfrac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})} \\ & =2 \end{aligned} $

Hence, the correct option is (B).

Solution

$ \begin{aligned} \sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}} & =\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1} \times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}} \\ & =\sqrt{\dfrac{(\sqrt{2}-1)^{2}}{(\sqrt{2})^{2}-1^{2}}} \\ & =\sqrt{\dfrac{(\sqrt{2}-1)^{2}}{2-1}} \\ & =\sqrt{\dfrac{(\sqrt{2}-1)^{2}}{1}} \\ & =1.4142-1 \\ & =0.4142 \end{aligned} $

Hence, the correct option is (C).

Solution

$ \begin{aligned} \sqrt[4]{\sqrt[3]{2^{2}}} & =\sqrt[4]{(2^{2})^{\dfrac{1}{3}}} \\ & =(2^{\dfrac{2}{3}})^{\dfrac{1}{4}} \\ & =2^{\dfrac{2}{3^{3}} \dfrac{1}{4}} \\ & =2^{\dfrac{1}{6}} \end{aligned} $

Hence, the correct option is (C).

Solution

$ \begin{aligned} \sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} & =2^{\dfrac{1}{2}} \times 2^{\dfrac{1}{4}} \times(2^{5})^{\dfrac{1}{12}} \\ & =2^{\dfrac{1}{3}} \times 2^{\dfrac{1}{4}} \times 2^{\dfrac{5}{12}} \\ & =2^{\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{5}{12}} \\ & =2^{\dfrac{4+3+5}{12}} \\ & =2^{\dfrac{12}{12}} \\ & =2 \end{aligned} $

Hence, the correct option is (B).

Solution

$ \begin{aligned} \sqrt[4]{(81)^{-2}} & =\sqrt[4]{(\dfrac{1}{81})^{2}} \\ & =(\dfrac{1}{81})^{2 \times \dfrac{1}{4}} \\ & =(\dfrac{1}{81})^{\dfrac{1}{2}} \\ & =\dfrac{1}{9} \end{aligned} $

Hence, the correct option is (A).

Solution

$ \begin{aligned} (256)^{0.16} \times(256)^{0.09} & =(256)^{0.16+0.09} \\ & =256^{0.25} 1 \\ & =256^{\dfrac{1}{4}} \\ & =4^{4 \times \dfrac{1}{4}} \\ & =4 \end{aligned} $

Hence, the correct option is (A).

Solution

(A) $x^{\dfrac{12}{7}}-x^{\dfrac{5}{7}} \neq x$

(B)

$ \begin{aligned} \sqrt[12]{(x^{4})^{\dfrac{1}{3}}} & =\sqrt[12]{x^{\dfrac{4}{3}}} \\ & =(x^{\dfrac{4}{3}})^{\dfrac{1}{12}} \\ & =x^{\dfrac{4}{3} \times \dfrac{1}{12}} \\ & =x^{\dfrac{1}{9}} \neq x \end{aligned} $

(C) $(\sqrt{x^{3}})^{\dfrac{2}{3}}=x^{\dfrac{3}{2} \times \dfrac{2}{3}}=x$

(D) $x^{\dfrac{12}{7}} \times x^{\dfrac{7}{12}}=x^{\dfrac{12}{7}+\dfrac{7}{12}}=x^{\dfrac{193}{84}} \neq x$

Hence, the correct option is (C).

Solution

True, $x+y$ is necessary an irrational number.

Let $x=6$ and $\sqrt{3}$.

Now, $x+y=6+\sqrt{3}=6+1.732 \ldots$ which is non-terminating and non-repeating.

Therefore, $x+y$ is an irrational number.

Solution

Let $x=0$ is a rational number and $y=\sqrt{3}$ is a irrational number. $x y=0 \times \sqrt{3}=0$ Which is an irrational number.

Therefore, $x y$ is not necessarily an irrational number.

Solution

(i) $\dfrac{\sqrt{2}}{3}$ is a rational number.

(ii) We know that, in between two integer there are infinitely many integer.

(iii) Rational number between 15 and 18 is finite.

(iv) There are number which can be written in the form $\dfrac{p}{q}, q \neq 0, p, q$ both are not integers.

(v) The square of an irrational number is always rational.

(vi) $\dfrac{\sqrt{12}}{\sqrt{3}}$ cab not be a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.

(vii) $\dfrac{\sqrt{15}}{\sqrt{3}}$ can be written in the form $\dfrac{p}{q}$, where $q \neq 0$ so it a rational number.

Solution

(i) $\sqrt{196}=\sqrt{14^{2}}=14$, which is a rational number.

(ii) $3 \sqrt{18}=9 \sqrt{2}$, which is an irrational number.

(iii) $\sqrt{\dfrac{9}{27}}=\dfrac{1}{\sqrt{3}}$, which is an irrational number.

(iv) $\dfrac{\sqrt{28}}{\sqrt{343}}=\dfrac{\sqrt{4}}{\sqrt{49}}=\dfrac{2}{7}$, which is a rational number.

(v) $-\sqrt{0.4}=-\dfrac{2}{\sqrt{10}}$, which is an irrational number.

(vi) $\dfrac{\sqrt{12}}{\sqrt{75}}=\sqrt{\dfrac{4}{25}}=\dfrac{2}{5}$, which is a rational number.

(vii) 0.5918 is terminating decimal, Therefore, it is a rational number.

(viii) $(1+\sqrt{5})-(4+\sqrt{5})=-3$, which is a rational number.

(ix) $10.124124 \ldots$ is a decimal expansion which is non-terminating but recurring. Hence, it is a rational number.

(x) $1.010010001 \ldots$ is a decimal expansion which is non-terminating but recurring. Hence, it is a rational number.

Solution

(i)

(ii)

$ \begin{aligned} & x^{2}=5 \\ & x= \pm \sqrt{5} \end{aligned} $

Which is an irrational number.

$y^{2}=9$

$y=\sqrt{9}$

$y= \pm 3$

Which is a rational number.

(iii)

$ \begin{aligned} & z^{2}=0.04 \\ & z= \pm \sqrt{0.04} \\ & z= \pm 0.2 \end{aligned} $

Which is a rational number.

(iv)

$u^{2}=\dfrac{17}{4}$

$u= \pm \sqrt{\dfrac{17}{4}}$

$u= \pm \dfrac{\sqrt{17}}{2}$

Where, $\sqrt{17}$ is not an integer. Which is an irrational number.

Solution

(i) $-1.1,-1,2$ and $-1,3$ are three rational numbers, which are lying between -1 and -2 .

(ii) $0.101,0,102,0.103$ are three rational number which are lying between 0.1 and 0.11 .

(iii) $\dfrac{5}{7}=\dfrac{5}{7} \times \dfrac{10}{10}=\dfrac{50}{70}$ and $\dfrac{6}{7}=\dfrac{6}{7} \times \dfrac{10}{10}=\dfrac{60}{70}$

$\dfrac{51}{70}, \dfrac{52}{70}, \dfrac{53}{70}$ are three rational numbers lying between $\dfrac{50}{70}$ and $\dfrac{60}{70}$. It mean that lying

between $\dfrac{5}{7}$ and $\dfrac{6}{7}$.

(iv)

$\dfrac{1}{4}=\dfrac{1}{4} \times \dfrac{20}{20}=\dfrac{20}{80}$ and $\dfrac{1}{5}=\dfrac{1}{5} \times \dfrac{16}{16}=\dfrac{16}{80}$

Now, $\sqrt{2} \times \sqrt{3} \times \dfrac{18}{80}(=\dfrac{9}{10}), \dfrac{19}{80}$ are three rational numbers lying between $\dfrac{1}{4}$ and $\dfrac{1}{5}$.

Solution

(i) A rational number between 2 and 3 is: $\dfrac{2+3}{2}=\dfrac{5}{2}=2.5$

(ii) 0.04 is rational number which lies between 0 and 0.1 .

(iii)

$\dfrac{1}{3}=\dfrac{1}{3} \times \dfrac{4}{4}=\dfrac{4}{12}$ and $\dfrac{1}{2}=\dfrac{1}{2} \times \dfrac{6}{6}=\dfrac{6}{12}$

$\dfrac{5}{12}$ is a rational number between $\dfrac{4}{12}$ and $\dfrac{6}{12}$. Which is also lying between $\dfrac{1}{3}$ and $\dfrac{1}{2}$.

Now, $\dfrac{1}{3}=0.33333$ and $\dfrac{1}{2}=0.5$.

Now, $0.414114111 \ldots$ is a non-terminating and non-recurring decimal.

Hence, $0.414114111 \ldots$ is an irrational number lying between $\dfrac{1}{3}$ and $\dfrac{1}{2}$.

(iv) $\dfrac{-2}{5}=-0.4$ and $\dfrac{1}{2}=0.5$

0 is rational number between -0.4 and 0.5 i.e., 0 is a rational number between $\dfrac{-2}{5}$ and $\dfrac{1}{2}$

Again, $0.131131113 \ldots$ is a non-terminating and non-recurring decimal which lies between -0.4 and 0.5 .

Hence, $0.131131113 \ldots$ is an irrational number lying between $\dfrac{-2}{5}$ and $\dfrac{1}{2}$.

(v) 0.151 is a rational number between 0.15 and 0.16 . Similarly, $0.153,0.157$, etc, are rational number lying between 0.15 and 0.16 .

0.151151115 is an irrational number between 0.15 and 0.16 .

(vi) $\sqrt{2}=1.4142135 \ldots$ and $\sqrt{3}=1.732050807$

Now, 1.5 which lies between $1.4142135 \ldots$ and $1.732050807 \ldots$

Since, 1.5 is a rational number between $\sqrt{2}$ and $\sqrt{3}$.

Now, $1.5755755557 \ldots$ is an irrational number lying between $\sqrt{2}$ and $\sqrt{3}$.

(vii) 3 is a rational number between 2.357 and 3.121 .

Again, 3.101101110 is an irrational number between 2.357 and 3.121.

(viii) 0.00011 is a rational number between 0.0001 and 0.001

Again, 0.0001131331333 is an irrational number between 0.0001 and 0.001 . (ix) 1 is a rational number between 0.484848 and 3.623623 .

Again, $1.909009000 \ldots$ is an irrational number lying between 0.484848 and 3.623623 .

(x) $\quad 6.3753$ is an rational number between 6.375289 and 6.375738 .

Again, $6.37541411411 \ldots$ is an irrational number lying between 6.375289 and 6.375738 .

Solution

Solution

Presentation of $\sqrt{5}$ on number line:

We can write 5 as the sum of the square of two natural numbers:

$5=1+4$

$=1^{2}+2^{2}$

On the number line, take $OA=2$ units.

Draw $BA=1$ unit, perpendicular to $OA$ join $OB$.

By Pythagoras theorem, $O B=\sqrt{5}$

Using a compass with center $O$ and radius $OB$, draw an arc which intersects the number line at a point $C$. Then, $C$ corresponds to $\sqrt{5}$.

Presentation of $\sqrt{10}$ on number line:

We can write 10 as the sum of the square of two natural numbers:

$10=1+9$

$=1^{2}+3^{2}$

On the number line, take $OA=3$ units.

Draw $BA=1$ unit, perpendicular to $OA$ join $OB$.

By Pythagoras theorem, $O B=\sqrt{10}$

Using a compass with center $O$ and radius $OB$, draw an arc which intersects the number line at a point $C$. Then, $C$ corresponds to $\sqrt{10}$.

Presentation of $\sqrt{17}$ on number line:

We can write 17 as the sum of the square of two natural numbers:

$10=1+16$

$ =1^{2}+4^{2} $

On the number line, take $OA=4$ units.

Draw $BA=1$ unit, perpendicular to $OA$ join $OB$.

By Pythagoras theorem, $O B=\sqrt{17}$

Using a compass with center $O$ and radius $OB$, draw an arc which intersects the number line at a point $C$. Then, $C$ corresponds to $\sqrt{17}$.

Solution

(i) Presentation of $\sqrt{4.5}$ on number line:

Mark the distance 4.5 units from a fixed point $A$ on a given line to obtain a point $B$ such that $AB=4.5$ units. From $B$, mark a distance of 1 units and mark the new points as $C$.

Find the mid-point of $AC$ and mark that points as $O$. Draw a semicircle with center $O$ and radius $OC$.

Draw a line perpendicular to $AC$ passing through $B$ and intersecting the semicircle at $D$. Then, $BD=\sqrt{4.5}$.

Now, draw an arc with center $B$ and $B$ radius $BD$, which intersects the number line in $E$.

Thus, E represent $\sqrt{4.5}$.

(ii) Presentation of $\sqrt{5.6}$ on number line:

Mark the distance 5.6 units from a fixed point $A$ on a given line to obtain a point $B$ such that $A B=5.6$ units. From B, mark a distance of 1 units and mark the new points as $C$.

Find the mid-point of $AC$ and mark that points as $O$. Draw a semicircle with center $O$ and radius $OC$.

Draw a line perpendicular to $AC$ passing through $B$ and intersecting the semicircle at $D$. Then, $BD=\sqrt{5.6}$.

Now, draw an arc with center B and B radius BD, which intersects the number line in E.

Thus, E represent $\sqrt{5.6}$.

(iii) Presentation of $\sqrt{8.1}$ on number line:

Mark the distance 8.1 units from a fixed point $A$ on a given line to obtain a point $B$ such that $AB=8.1$ units. From B, mark a distance of 1 units and mark the new points as $C$.

Find the mid-point of $AC$ and mark that points as $O$. Draw a semicircle with center $O$ and radius $OC$.

Draw a line perpendicular to $AC$ passing through $B$ and intersecting the semicircle at $D$. Then, $BD=\sqrt{8.1}$.

Now, draw an arc with center B and B radius BD, which intersects the number line in $E$.

Thus, E represent $\sqrt{8.1}$.

(iv) Presentation of $\sqrt{2.3}$ on number line:

Mark the distance 2.3 units from a fixed point $A$ on a given line to obtain a point $B$ such that $AB=2.3$ units. From B, mark a distance of 1 units and mark the new points as $C$.

Find the mid-point of $AC$ and mark that points as $O$. Draw a semicircle with center $O$ and radius $OC$.

Draw a line perpendicular to $AC$ passing through $B$ and intersecting the semicircle at $D$. Then, $BD=\sqrt{2.3}$.

Now, draw an arc with center B and B radius BD, which intersects the number line in E.

Thus, E represent $\sqrt{2.3}$.

Solution

(i) $0.2=\dfrac{2}{10}=\dfrac{1}{5}$

(ii) Let $x=0.888 \ldots=0 . \overline{8}$

$10 x=8 . \overline{8}$

Subtracting (I) from (II), get:

$9 x=8$

Therefore, $x=\dfrac{8}{9}$.

(iii) Let $x=5 . \overline{2}=5.2222 \ldots$

Multiplying both sides by 10 , get:

$10 x=52.222 \ldots=52 . \overline{2}$

Subtracting (I) from (II), get:

$10 x-x=47$

$ 9 x=47 $

$ x=\dfrac{47}{9} $

Hence, $5 . \overline{2}=\dfrac{47}{9}$.

(iv) Let $x=0 . \overline{001}=0.001001$

$1000 x=1.001001 \ldots$

Subtracting (I) from (II), get:

$999 x=1$

Hence, $x=\dfrac{1}{999}$.

(v) Let $x=0.2555 \ldots=0.2 \overline{5}$. So,

$10 x=2 . \overline{5}$

And:

$100 x=25 . \overline{5}$

Subtracting (II) from (III), get:

$90 x=23$

$x=\dfrac{23}{90}$

(vi) Let $x=0.1 \overline{34}=0.1343434$

Multiplying both sides by 100 , get:

$100 x=13.43434=13.4 \overline{34}$

Subtracting (I) from (II), get:

Solution

Consider the expression:

$0.6+0 . \overline{7}+0.4 \overline{7}$

We have:

$0.6=\dfrac{6}{10}$

Let

$x=0 . \overline{7}=0.777 \ldots$

And: $10 x=7.77 \ldots$

Subtract equation (I) from equation (II), get:

$9 x=7$

$x=\dfrac{7}{9}$

Similarly: Let $y=0.4 \overline{7}=0.4777 \ldots$

Now, $10 y=4 . \overline{7}$

$100 y=47 . \overline{7}$

Subtract equation (III) from equation (IV), get:

$90 y=43$

$ y=\dfrac{43}{90} $

$0.4 \overline{7}=\dfrac{43}{90}$

Now,

$ \begin{aligned} 0.6+0 . \overline{7}+0.4 \overline{7} & =\dfrac{6}{10}+\dfrac{7}{9}+\dfrac{43}{90} \\ & =\dfrac{54+70+43}{90} \\ & =\dfrac{167}{90} \end{aligned} $

Therefore, $\dfrac{167}{90}$ in the form $\dfrac{p}{q}$ and $q \neq 0$.

Solution

Consider the expression: $ \begin{aligned}\dfrac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\dfrac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\dfrac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}\end{aligned} $

Simplify the above expression as follows:

$\begin{aligned} \dfrac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\dfrac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\dfrac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}} \\ =\dfrac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}} \times \dfrac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}-\dfrac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}} \times \dfrac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\ - \dfrac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}} \times \dfrac{\sqrt{15}-3 \sqrt{2}}{\sqrt{15}-3 \sqrt{2}} \\ = \dfrac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{(\sqrt{10})^{2}-(\sqrt{3})^{2}}-\dfrac{2 \sqrt{5}(\sqrt{6}-\sqrt{5})}{(\sqrt{6})^{2}-(\sqrt{5})^{2}}-\dfrac{3 \sqrt{2}(\sqrt{15}-3 \sqrt{2})}{(\sqrt{15})^{2}-(3 \sqrt{2})^{2}} \\ = \dfrac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{10-3}-\dfrac{2 \sqrt{5}(\sqrt{6}-\sqrt{5})}{6-5}-\dfrac{3 \sqrt{2}(\sqrt{15}-3 \sqrt{2})}{15-18} \\ = \dfrac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{7}-\dfrac{2 \sqrt{5}(\sqrt{6}-\sqrt{5})}{1}-\dfrac{3 \sqrt{2}(\sqrt{15}-3 \sqrt{2})}{-3} \\ = \sqrt{3}(\sqrt{10}-\sqrt{3})-2 \sqrt{5}(\sqrt{6}-\sqrt{5})+\sqrt{2}(\sqrt{15}-3 \sqrt{2}) \\ = \sqrt{30}-3-2 \sqrt{30}+10+\sqrt{30}-6 \\ =2\sqrt{30}-9-2\sqrt{30}+10=1 \end{aligned}$

Solution

Consider the expression:

$\dfrac{4}{3 \sqrt{3}-2 \sqrt{2}}+\dfrac{3}{3 \sqrt{3}+2 \sqrt{2}}$

Rationalization the above expression as follows:

$ \begin{aligned} \dfrac{4}{3 \sqrt{3}-2 \sqrt{2}}+\dfrac{3}{3 \sqrt{3}+2 \sqrt{2}} & =\dfrac{4}{3 \sqrt{3}-2 \sqrt{2}} \times \dfrac{3 \sqrt{3}+2 \sqrt{2}}{3 \sqrt{3}+2 \sqrt{2}}+\dfrac{3}{3 \sqrt{3}+2 \sqrt{2}} \\ \times \dfrac{3 \sqrt{3}-2 \sqrt{2}}{3 \sqrt{3}-2 \sqrt{2}} \\ & =\dfrac{4(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}}+\dfrac{3(3 \sqrt{3}-2 \sqrt{2})}{(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}} \\ & =\dfrac{4(3 \sqrt{3}+2 \sqrt{2})}{27-8}+\dfrac{3(3 \sqrt{3}-2 \sqrt{2})}{27-8} \\ & =\dfrac{12 \sqrt{3}+8 \sqrt{2}+9 \sqrt{3}-6 \sqrt{2}}{27-8} \\ & =\dfrac{12 \sqrt{3}+8 \sqrt{2}+9 \sqrt{3}-6 \sqrt{2}}{19} \\ & =\dfrac{21 \sqrt{3}+2 \sqrt{2}}{19} \end{aligned} $

Substitute 1.414 for $\sqrt{2}$ and 1.732 for $\sqrt{3}$ in the above expression.

$\dfrac{21 \times 1.732+2 \times 1.414}{19}=2.063$

Solution

Given:

$a=\dfrac{3+\sqrt{5}}{2}$

The value of $a^{2}$ will be:

$ \begin{aligned} a^{2} & =(\dfrac{3+\sqrt{5}}{2})^{2} \\ & =\dfrac{9+5+6 \sqrt{5}}{4} \\ & =\dfrac{14+6 \sqrt{5}}{4} \\ & =\dfrac{7+3 \sqrt{5}}{2} \end{aligned} $

Now,

$ \begin{aligned} \dfrac{1}{a^{2}} & =\dfrac{2}{7+3 \sqrt{5}} \\ & =\dfrac{2}{7+3 \sqrt{5}} \times \dfrac{7-3 \sqrt{5}}{7-3 \sqrt{5}} \\ & =\dfrac{2(7-3 \sqrt{5})}{7^{2}-(3 \sqrt{5})^{2}} \\ & =\dfrac{2(7-3 \sqrt{5})}{49-45} \\ & =\dfrac{2(7-3 \sqrt{5})}{4} \\ & =\dfrac{7-3 \sqrt{5}}{2} \end{aligned} $

The value of $a^{2}+\dfrac{1}{a^{2}}$ is:

$ \begin{aligned} a^{2}+\dfrac{1}{a^{2}} & =\dfrac{7+3 \sqrt{5}}{2}+\dfrac{7-3 \sqrt{5}}{2} \\ & =\dfrac{7+3 \sqrt{5}+7-3 \sqrt{5}}{2} \\ & =\dfrac{14}{2} \\ & =7 \end{aligned} $

Solution

Given:

$ x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \text{ and } y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} $

Rationalization the $x$ as follows:

$ \begin{aligned} x & =\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ & =\dfrac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}} \end{aligned} $

$ \begin{aligned} & =\dfrac{(\sqrt{3})^{2}+(\sqrt{2})^{2}+2 \times \sqrt{3} \times \sqrt{2}}{3-2} \\ & =\dfrac{3+2+2 \times \sqrt{6}}{1} \\ & =5+2 \sqrt{6} \end{aligned} $

Similarly: $y=5-2 \sqrt{6}$

Now,

$ \begin{aligned} x+y & =5+2 \sqrt{6}+5-2 \sqrt{6} \\ & =10 \end{aligned} $

And,

$ x y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} $

$ =1 $

Therefore,

$ \begin{aligned} x+y & =(10)^{2}-(1)^{2} \\ & =100-1 \\ & =99 \end{aligned} $

Solution

Consider the expression: $(256)^{-(\dfrac{-3}{4^{2}})}$

Now, simplify the above expression as follows:

$ \begin{aligned} (256)^{-(\dfrac{-3}{4^{2}})} & =2^{8-(4^{-\dfrac{3}{2}})} \\ & =2^{8-(2^{2 \times \dfrac{3}{2}})} \\ & =(2^{8})^{-(2^{-3})} \\ & =(2^{8})^{-\dfrac{1}{8}} \\ & =2^{8 \times-\dfrac{1}{8}} \\ & =2^{-1} \\ & =\dfrac{1}{2} \end{aligned} $

Solution

Consider the expression:

$\dfrac{4}{(216)^{-\dfrac{2}{3}}}+\dfrac{1}{(256)^{-\dfrac{3}{4}}}+\dfrac{2}{(243)^{-\dfrac{1}{5}}}$

Simplify the above expression as follows:

$ \begin{aligned} \dfrac{4}{(216)^{-\dfrac{2}{3}}}+\dfrac{1}{(256)^{-\dfrac{3}{4}}}+\dfrac{2}{(243)^{-\dfrac{1}{5}}} & 4 \times(216)^{\dfrac{2}{3}}+(256)^{\dfrac{3}{4}}+2 \times(243)^{\dfrac{1}{5}} \\ & =4 \times(216)^{\dfrac{2}{3}}+(256)^{\dfrac{3}{4}}+2 \times(243)^{\dfrac{1}{5}} \\ & =4 \times 6^{3 \times \dfrac{2}{3}}+4^{4 \times \dfrac{3}{4}}+2 \times 3^{5 \times \dfrac{1}{5}} \\ & =4 \times 6^{2}+4^{3}+2 \times 3 \\ & =144+64+6 \\ & =214 \end{aligned} $