Question

best example.. question kya and answer kya.. (눈‸눈).. shi h .. krlo pareshaan​

Answer 1

lol..

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Explanation:

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$\displaystyle\sf x = 3+2\sqrt{2}$

$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$

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$\begin{gathered}\displaystyle\sf :\implies \dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}}\\\end{gathered}$

$\displaystyle\sf :\implies \dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{3^2-(2\sqrt{2}^2)}$

$\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{9-8}$

$\displaystyle\sf :\implies \dfrac{1}{x} = 3-2\sqrt{2}$

__________________

$\displaystyle\sf :\implies x+\dfrac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2})$

$\displaystyle\sf :\implies x+\dfrac{1}{x} = 3+2\sqrt{2} + 3 - 2\sqrt{2}$

$\displaystyle\sf :\implies x+\dfrac{1}{x}$

So here we know that we may split the number 6 into 4+2 and 4+2 = 6

$\displaystyle\sf :\implies x+\dfrac{1}{x} = 4$

$\displaystyle\sf :\implies x+\dfrac{1}{x}-2 = 4$

$\displaystyle\sf :\implies \bigg\lgroup \sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg\rgroup^2 = 4$

$\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \sqrt{4}$

$\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \pm 2$

$\displaystyle\therefore\:\underline{\textsf{The value of \sqrt{ \sf x}-\dfrac{\sf 1}{\sqrt{\sf x}} is \textbf{ \pm 2 }}}$

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Answer 2

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\displaystyle\sf x = 3+2\sqrt{2}x=3+2

2

\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}

x

−

x

1

━━━━━━━━

\begin{gathered} \begin{gathered}\displaystyle\sf :\implies \dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}}\\\end{gathered}\end{gathered}

:⟹

x

1

=

3+2

2

1

\displaystyle\sf :\implies \dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}:⟹

3+2

2

1

×

3−2

2

3−2

2

\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{3^2-(2\sqrt{2}^2)}:⟹

3

2

−(2

2

2

)

3−2

2

\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{9-8}:⟹

9−8

3−2

2

\displaystyle\sf :\implies \dfrac{1}{x} = 3-2\sqrt{2}:⟹

x

1

=3−2

2

__________________

\displaystyle\sf :\implies x+\dfrac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2}):⟹x+

x

1

=(3+2

2

)+(3−2

2

)

\displaystyle\sf :\implies x+\dfrac{1}{x} = 3+2\sqrt{2} + 3 - 2\sqrt{2}:⟹x+

x

1

=3+2

2

+3−2

2

\displaystyle\sf :\implies x+\dfrac{1}{x}:⟹x+

x

1

So here we know that we may split the number 6 into 4+2 and 4+2 = 6

\displaystyle\sf :\implies x+\dfrac{1}{x} = 4:⟹x+

x

1

=4

\displaystyle\sf :\implies x+\dfrac{1}{x}-2 = 4:⟹x+

x

1

−2=4

\displaystyle\sf :\implies \bigg\lgroup \sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg\rgroup^2 = 4:⟹

⎩

⎧

x

−

x

1

⎭

⎫

2

=4

\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \sqrt{4}:⟹

x

−

x

1

=

4

\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \pm 2:⟹

x

−

x

1

=±2

\displaystyle\therefore\:\underline{\textsf{The value of $ \sqrt{ \sf x}-\dfrac{\sf 1}{\sqrt{\sf x}}$ is \textbf{$\pm$2 }}}∴

The value of

x

−

x

1

is ±2

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