Question

x=5-√21 then the value of √x÷(√(32-2x-√21))

Answer 1

$Given, \\ x=5- \sqrt{21} \\ = \frac{\sqrt{5-\sqrt{21}}}{ \sqrt{32-10+2\sqrt{21}-\sqrt{21}}} \\ \frac{\sqrt{5-\sqrt{21}}}{\sqrt{21}+1}$

Answer 2

The value of $\dfrac{\sqrt{x}}{\sqrt{32-2x-\sqrt{21}}}=\dfrac{\sqrt{5-\sqrt{21}}}{\sqrt{22+\sqrt{21}}}$.

Step-by-step explanation:

We have,

$x=5-\sqrt{21}$

To find, the value of $\dfrac{\sqrt{x}}{\sqrt{32-2x-\sqrt{21}}}$=?

∴  $\dfrac{\sqrt{x}}{\sqrt{32-2x-\sqrt{21}}}$

Put $x=5-\sqrt{21}$, we get

$=\dfrac{\sqrt{5-\sqrt{21}}}{\sqrt{32-2(5-\sqrt{21})-\sqrt{21}}}$

$=\dfrac{\sqrt{5-\sqrt{21}}}{\sqrt{32-10+2\sqrt{21}-\sqrt{21}}}$

$=\dfrac{\sqrt{5-\sqrt{21}}}{\sqrt{32-10+\sqrt{21}}}$

$=\dfrac{\sqrt{5-\sqrt{21}}}{\sqrt{22+\sqrt{21}}}$

∴ The value of $\dfrac{\sqrt{x}}{\sqrt{32-2x-\sqrt{21}}}=\dfrac{\sqrt{5-\sqrt{21}}}{\sqrt{22+\sqrt{21}}}$

Hence, the value of $\dfrac{\sqrt{x}}{\sqrt{32-2x-\sqrt{21}}}=\dfrac{\sqrt{5-\sqrt{21}}}{\sqrt{22+\sqrt{21}}}$.