Study91

On simplification the value of $1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} is = ?$ ${\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}$

1 $2 \sqrt{2} - 1$ $2\sqrt 2 - 1$
2 $1 - 2 \sqrt{2}$ $1 - 2\sqrt 2$
3 $1 - \sqrt{2}$ $1 - \sqrt 2$
4 $- 2 \sqrt{2}$ $- 2\sqrt 2$

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} \\ = 1 - \frac{(\sqrt{2} - 1)}{(\sqrt{2} + 1) (\sqrt{2} - 1)} - \frac{(\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)} \\ = 1 - \sqrt{2} + 1 - \sqrt{2} - 1 \\ = 1 - 2 \sqrt{2} \end{aligned}

$\eqalign{ & {\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{ }} \cr & = 1 - \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} - \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} \cr & = 1 - \sqrt 2 + 1 - \sqrt 2 - 1 \cr & = 1 - 2\sqrt 2 \cr}$

If $(x - \frac{1}{x}) = \sqrt{21},$ $\left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} {\text{,}}$ then the value of $(x^{2} + \frac{1}{x^{2}}) (x + \frac{1}{x})$ $\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {x + \frac{1}{x}} \right)$ is = ?

142
263
3115
4120

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (x - \frac{1}{x}) = \sqrt{21} \\ \Rightarrow {(x - \frac{1}{x})}^{2} = {(\sqrt{21})}^{2} \\ = 21 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} - 2 = 21 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = 23 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 25 \\ \Rightarrow {(x + \frac{1}{x})}^{2} = 5^{2} \\ \Rightarrow x + \frac{1}{x} = 5 \\ ∴ (x^{2} + \frac{1}{x^{2}}) (x + \frac{1}{x}) \\ = 23 \times 5 \\ = 115 \end{aligned}

$\eqalign{ & \left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = }}{\left( {\sqrt {21} } \right)^2} \cr & \,\,\,\,\,\,\, = 21 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 21 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 23 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 25 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{5}}^2} \cr & \Rightarrow x + \frac{1}{x} = 5 \cr & \therefore \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {x + \frac{1}{x}} \right) \cr & = 23 \times 5 \cr & = 115 \cr}$

If $(a + \frac{1}{a}) = 6,$ $\left( {a + \frac{1}{a}} \right) = 6,$ then $(a^{4} + \frac{1}{a^{4}})$ $\left( {{a^4} + \frac{1}{{{a^4}}}} \right)$ = ?

11154
21158
31160
41164

View Answer Discuss in forum Workspace Report

Answer:- 4

Explanation:-

Solution:

\begin{aligned} (a + \frac{1}{a}) = 6 \\ \Rightarrow {(a + \frac{1}{a})}^{2} = 6^{2} \\ = 36 \\ \Rightarrow a^{2} + \frac{1}{a^{2}} + 2 = 36 \\ \Rightarrow (a^{2} + \frac{1}{a^{2}}) = 34 \\ \Rightarrow {(a^{2} + \frac{1}{a^{2}})}^{2} = {(34)}^{2} \\ \Rightarrow a^{4} + \frac{1}{a^{4}} + 2 = 1156 \\ \Rightarrow (a^{4} + \frac{1}{a^{4}}) = 1154. \end{aligned}

$\eqalign{ & \left( {a + \frac{1}{a}} \right) = 6 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^2} = {6^2} \cr & \,\,\,\,\,\,\,\, = 36 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 36 \cr & \Rightarrow \left( {{a^2} + \frac{1}{{{a^2}}}} \right) = 34 \cr & \Rightarrow {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} = {\left( {34} \right)^2} \cr & \Rightarrow {a^4} + \frac{1}{{{a^4}}} + 2 = 1156 \cr & \Rightarrow \left( {{a^4} + \frac{1}{{{a^4}}}} \right) = 1154. \cr}$

The simplest value of $(\frac{1}{\sqrt{9} - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}}) (\frac{1}{\sqrt{9} - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}}) = ?$ $\left( {\frac{1}{{\sqrt 9 - \sqrt 8 }} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }}} \right)\left( {\frac{1}{{\sqrt 9 - \sqrt 8 }} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }}} \right) = ?$

1 $3 - \sqrt{5}$ $3 - \sqrt 5$
23
3 $\sqrt{5}$ $\sqrt 5$
4 $\sqrt{5} - 3$ $\sqrt 5 - 3$

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (\frac{1}{\sqrt{9} - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}}) \\ = \sqrt{9} + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} \\ = \sqrt{9} - \sqrt{5} \\ = 3 - \sqrt{5} \end{aligned}

$\eqalign{ & \left( {\frac{1}{{\sqrt 9 - \sqrt 8 }} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }}} \right) \cr & = \sqrt 9 + \sqrt 8 - \sqrt 8 - \sqrt 7 + \sqrt 7 + \sqrt 6 - \sqrt 6 - \sqrt 5 \cr & = \sqrt 9 - \sqrt 5 \cr & = 3 - \sqrt 5 \cr}$

If $(x + \frac{1}{x}) = 2,$ $\left( {x + \frac{1}{x}} \right){\text{ = 2,}}$ then $(x - \frac{1}{x})$ $\left( {x - \frac{1}{x}} \right)$ is equal to = ?

10
21
32
45

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (x + \frac{1}{x}) = 2 \\ \Rightarrow {(x + \frac{1}{x})}^{2} = 2^{2} \\ \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 4 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = 2 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} - 2. x . \frac{1}{x} \\ = 2 - 2 = 0 \\ \Rightarrow {(x - \frac{1}{x})}^{2} =  0 \\ \Rightarrow x - \frac{1}{x} = 0 \end{aligned}

$\eqalign{ & \left( {x + \frac{1}{x}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{2}}^2} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 4 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} \cr & \,\,\,\,\,\,\,\, = 2 - 2 = 0 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = 0}} \cr & \Rightarrow x - \frac{1}{x} = 0 \cr}$

$(x + \frac{1}{x}) (x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - 1) (x^{2} + \frac{1}{x^{2}} + 1)$ $\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$ is equal to ?

1 $x^{6} - \frac{1}{x^{6}}$ ${x^6} - \frac{1}{{{x^6}}}$
2 $x^{8} - \frac{1}{x^{8}}$ ${x^8} - \frac{1}{{{x^8}}}$
3 $x^{6} + \frac{1}{x^{6}}$ ${x^6} + \frac{1}{{{x^6}}}$
4 $x^{8} + \frac{1}{x^{8}}$ ${x^8} + \frac{1}{{{x^8}}}$

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression , \\ [(x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - x . \frac{1}{x})] [(x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} x . \frac{1}{x})] \\ = (x^{3} + \frac{1}{x^{3}}) (x^{3} + \frac{1}{x^{3}}) \\ = x^{6} + \frac{1}{x^{6}} \end{aligned}

$\eqalign{ & {\text{Given expression ,}} \cr & \,\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}x.\frac{1}{x}} \right)} \right] \cr & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = {x^6} + \frac{1}{{{x^6}}} \cr}$

The expression $\frac{1}{x - 1} - \frac{1}{x + 1} - \frac{2}{x^{2} + 1} - \frac{4}{x^{4} + 1}$ $\frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}}$ is equal to = ?

1 $\frac{8}{x^{8} + 1}$ $\frac{8}{{{x^8} + 1}}$
2 $\frac{8}{x^{8} - 1}$ $\frac{8}{{{x^8} - 1}}$
3 $\frac{8}{x^{7} - 1}$ $\frac{8}{{{x^7} - 1}}$
4 $\frac{8}{x^{7} + 1}$ $\frac{8}{{{x^7} + 1}}$

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression, \\ (\frac{1}{x - 1} - \frac{1}{x + 1}) - \frac{2}{x^{2} + 1} - \frac{4}{x^{4} + 1} \\ = [\frac{(x + 1) - (x - 1)}{(x - 1) (x + 1)}] - \frac{2}{x^{2} + 1} - \frac{4}{x^{4} + 1} \\ = (\frac{2}{x^{2} - 1} - \frac{2}{x^{2} + 1}) - \frac{4}{x^{4} + 1} \\ = [\frac{2 (x^{2} + 1) - (x^{2} - 1)}{(x^{2} - 1) (x^{2} + 1)}] - \frac{4}{x^{4} + 1} \\ = \frac{4}{x^{4} - 1} - \frac{4}{x^{4} + 1} \\ = \frac{4 (x^{4} + 1) - 4 (x^{4} - 1)}{(x^{4} - 1) (x^{4} + 1)} \\ = \frac{8}{x^{8} - 1} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} \cr & = \left[ {\frac{{\left( {x + 1} \right) - \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right] - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} \cr & = \left( {\frac{2}{{{x^2} - 1}} - \frac{2}{{{x^2} + 1}}} \right) - \frac{4}{{{x^4} + 1}} \cr & = \left[ {\frac{{2\left( {{x^2} + 1} \right) - \left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}} \right] - \frac{4}{{{x^4} + 1}} \cr & = \frac{4}{{{x^4} - 1}} - \frac{4}{{{x^4} + 1}} \cr & = \frac{{4\left( {{x^4} + 1} \right) - 4\left( {{x^4} - 1} \right)}}{{\left( {{x^4} - 1} \right)\left( {{x^4} + 1} \right)}} \cr & = \frac{8}{{{x^8} - 1}} \cr}$

08. The Value of (√6 + √10 - √21 - √35) × (√6 - √10 + √21 - √35) = ?

127
218
340
410

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

(√6 + √10 - √21 - √35) × (√6 - √10 + √21 - √35)
= {(√6 - √35) + (√10 - √21)} × {(√6 - √35) - (√10 - √21)}
= (√6 - √35)² - (√10 - √21)²
= 6 + 35 - 2√210 - 10 - 21 + 2√210
= 41 - 31
= 10

If $x = \sqrt{3} + \sqrt{2},$ $x = \sqrt 3 {\text{ + }}\sqrt 2 {\text{,}}$ then the value of $x^{3} - \frac{1}{x^{3}}$ ${x^3} - \frac{1}{{{x^3}}}$ is?

1 $10 \sqrt{2}$ $10\sqrt 2$
2 $14 \sqrt{2}$ $14\sqrt 2$
3 $22 \sqrt{2}$ $22\sqrt 2$
4 $8 \sqrt{2}$ $8\sqrt 2$

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \sqrt{3} + \sqrt{2} \\ \frac{1}{x} = \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\ \frac{1}{x} = \sqrt{3} - \sqrt{2} \\ x^{3} - \frac{1}{x^{3}} \\ = {[x - \frac{1}{x}]}^{3} + 3 \times x \times \frac{1}{x} (x - \frac{1}{x}) \\ = {(\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2})}^{3} + 3 (\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}) \\ = {(2 \sqrt{2})}^{3} + 3 (2 \sqrt{2}) \\ = 16 \sqrt{2} + 6 \sqrt{2} \\ = 22 \sqrt{2} \end{aligned}

$\eqalign{ & x = \sqrt 3 + \sqrt 2 \cr & \frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \frac{1}{x} = \sqrt 3 - \sqrt 2 \cr & {x^3} - \frac{1}{{{x^3}}} \cr & = {\left[ {x - \frac{1}{x}} \right]^3} + 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) \cr & = {\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)^3} + 3\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right) \cr & = {\left( {2\sqrt 2 } \right)^3} + 3\left( {2\sqrt 2 } \right) \cr & = 16\sqrt 2 + 6\sqrt 2 \cr & = 22\sqrt 2 \cr}$

If $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$ $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$ and $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$ $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$ where a, b, c, p, q, r are non-zero real numbers, then $\frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}}$ $\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}$ is equal to = ?

10
21
33
49

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0 \\ \Rightarrow a q r + b p r + c p q = 0.... (i) \\ \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1 \\ \Rightarrow {(\frac{p}{a} + \frac{q}{b} + \frac{r}{c})}^{2} = 1 \\ \Rightarrow \frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}} + 2 (\frac{p q}{a b} + \frac{p r}{a c} + \frac{q r}{b c}) = 1 \\ \Rightarrow \frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}} + \frac{2 (p q c + p r b + q r a)}{a b c} = 1 \\ \Rightarrow \frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}} = 1.... [using (i)] \end{aligned}

$\eqalign{ & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0\,\,\,\, \cr & \Rightarrow aqr + bpr + cpq = 0....({\text{i}}) \cr & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1\,\, \cr & \Rightarrow {\left( {\frac{p}{a} + \frac{q}{b} + \frac{r}{c}} \right)^2} = 1 \cr & \Rightarrow {\text{ }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}\, + 2\left( {\frac{{pq}}{{ab}} + \frac{{pr}}{{ac}} + \frac{{qr}}{{bc}}} \right) = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} + \frac{{2\left( {pqc + prb + qra} \right)}}{{abc}} = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1....\left[ {{\text{using (i)}}} \right] \cr}$

Hello Guest

Simplification - 02

On simplification the value of $1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} is = ?$ ${\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}$

If $(x - \frac{1}{x}) = \sqrt{21},$ $\left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} {\text{,}}$ then the value of $(x^{2} + \frac{1}{x^{2}}) (x + \frac{1}{x})$ $\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {x + \frac{1}{x}} \right)$ is = ?

If $(a + \frac{1}{a}) = 6,$ $\left( {a + \frac{1}{a}} \right) = 6,$ then $(a^{4} + \frac{1}{a^{4}})$ $\left( {{a^4} + \frac{1}{{{a^4}}}} \right)$ = ?

If $(x + \frac{1}{x}) = 2,$ $\left( {x + \frac{1}{x}} \right){\text{ = 2,}}$ then $(x - \frac{1}{x})$ $\left( {x - \frac{1}{x}} \right)$ is equal to = ?

$(x + \frac{1}{x}) (x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - 1) (x^{2} + \frac{1}{x^{2}} + 1)$ $\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$ is equal to ?

The expression $\frac{1}{x - 1} - \frac{1}{x + 1} - \frac{2}{x^{2} + 1} - \frac{4}{x^{4} + 1}$ $\frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}}$ is equal to = ?

08. The Value of (√6 + √10 - √21 - √35) × (√6 - √10 + √21 - √35) = ?

If $x = \sqrt{3} + \sqrt{2},$ $x = \sqrt 3 {\text{ + }}\sqrt 2 {\text{,}}$ then the value of $x^{3} - \frac{1}{x^{3}}$ ${x^3} - \frac{1}{{{x^3}}}$ is?

Subject Wise

Exam Wise

NCERT
6 to 12

Paid Course

Practice Batch

Daily Current Affairs

Monthly Current Affairs

Topic Wise Current Affairs

State Wise Current Affairs

Prelims Facts

Hello Guest

Simplification - 02

01. On simplification the value of 1−11+2–√ + 11−2–√is = ?1−11+2 + 11−2is = ?{\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}

02. If (x−1x) = 21−−√,(x−1x) = 21,\left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} {\text{,}} then the value of (x2+1x2)(x+1x)(x2+1x2)(x+1x)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {x + \frac{1}{x}} \right) is = ?

03. If (a+1a)=6,(a+1a)=6,\left( {a + \frac{1}{a}} \right) = 6, then (a4+1a4)(a4+1a4)\left( {{a^4} + \frac{1}{{{a^4}}}} \right) = ?

05. If (x+1x) = 2,(x+1x) = 2,\left( {x + \frac{1}{x}} \right){\text{ = 2,}} then (x−1x)(x−1x)\left( {x - \frac{1}{x}} \right) is equal to = ?

06. (x+1x)(x−1x)(x2+1x2−1)(x2+1x2+1)(x+1x)(x−1x)(x2+1x2−1)(x2+1x2+1)\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right) is equal to ?

07. The expression 1x−1−1x+1−2x2+1−4x4+11x−1−1x+1−2x2+1−4x4+1\frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} is equal to = ?

08. The Value of (√6 + √10 - √21 - √35) × (√6 - √10 + √21 - √35) = ?

09. If x=3–√ + 2–√,x=3 + 2,x = \sqrt 3 {\text{ + }}\sqrt 2 {\text{,}} then the value of x3−1x3x3−1x3{x^3} - \frac{1}{{{x^3}}} is?

Subject Wise

Exam Wise

NCERT 6 to 12

Paid Course

Practice Batch

On simplification the value of $1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} is = ?$ ${\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}$

If $(x - \frac{1}{x}) = \sqrt{21},$ $\left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} {\text{,}}$ then the value of $(x^{2} + \frac{1}{x^{2}}) (x + \frac{1}{x})$ $\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {x + \frac{1}{x}} \right)$ is = ?

If $(a + \frac{1}{a}) = 6,$ $\left( {a + \frac{1}{a}} \right) = 6,$ then $(a^{4} + \frac{1}{a^{4}})$ $\left( {{a^4} + \frac{1}{{{a^4}}}} \right)$ = ?

If $(x + \frac{1}{x}) = 2,$ $\left( {x + \frac{1}{x}} \right){\text{ = 2,}}$ then $(x - \frac{1}{x})$ $\left( {x - \frac{1}{x}} \right)$ is equal to = ?

$(x + \frac{1}{x}) (x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - 1) (x^{2} + \frac{1}{x^{2}} + 1)$ $\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$ is equal to ?

The expression $\frac{1}{x - 1} - \frac{1}{x + 1} - \frac{2}{x^{2} + 1} - \frac{4}{x^{4} + 1}$ $\frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}}$ is equal to = ?

If $x = \sqrt{3} + \sqrt{2},$ $x = \sqrt 3 {\text{ + }}\sqrt 2 {\text{,}}$ then the value of $x^{3} - \frac{1}{x^{3}}$ ${x^3} - \frac{1}{{{x^3}}}$ is?

NCERT
6 to 12