Question

i have doubt in this question please solve this question

Answer 1

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it's right

Answer 2

Given ---
x = 3 + 2√2

To find ---
√x - 1/√x

Finding ---

$x = 3 + 2 \sqrt{2}$
$= > \sqrt{x} = \sqrt{3 + 2 \sqrt{2} }$
$= > \sqrt{x} = \sqrt{( { \sqrt{2}) }^{2} + ( {1})^{2} + (2 \times 1 \times \sqrt{2} ) }$
$= > \sqrt{x } = \sqrt{ ({ \sqrt{2} + 1})^{2} }$
$= > \sqrt{x} = \sqrt{2} + 1$
Therefore, √x = √2 + 1

Now,
$\frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{2} + 1 }$
$= > \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{2 } + 1} \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1 }$
$= > \frac{1}{ \sqrt{x} } = \frac{1(2 - \sqrt{1}) }{ ({ \sqrt{2}) }^{2} - ( {1})^{2} }$
$= > \frac{1}{ \sqrt{x} } = \frac{2 - \sqrt{1} }{2 - 1}$
$= > \frac{1}{ \sqrt{x} } = \frac{2 - \sqrt{1} }{1}$
$\frac{1}{ \sqrt{x} } = 2 - \sqrt{1}$
Therefore, 1/√x = 2 - √1

Now,

$\sqrt{x} - \frac{1}{ \sqrt{x} } = (2 + \sqrt{1} ) - (2 - \sqrt{1} )$
$= > \sqrt{x} - \frac{1}{ \sqrt{x} } = 2 + \sqrt{1} - 2 + \sqrt{1}$
$\sqrt{x} - \frac{1}{ \sqrt{x} } = 2 \sqrt{2}$
That's your answer.

Hope it'll help.. :-D