Let y=2+2,z=2−2 and x=y+z, then x is

# Let $y=\sqrt{2+\sqrt{2}},z=\sqrt{2-\sqrt{2}}$ and $x=y+z$, then $x$ is

1. A

$\sqrt{2+\sqrt{2}}$

2. B

$\sqrt{\sqrt{2}-\sqrt{2+\sqrt{2}}}$

3. C

$\sqrt{2}.\sqrt{2+\sqrt{2}}$

4. D

$\sqrt{2}.\sqrt{2-\sqrt{2}}$

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### Solution:

Given $y=\sqrt{2+\sqrt{2}},z=\sqrt{2-\sqrt{2}}$

$\therefore x=y+z=\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$

${x}^{2}=\left(2+\sqrt{2}\right)+\left(2-\sqrt{2}\right)+2\sqrt{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}$

${x}^{2}=4+2\sqrt{2}$

${x}^{2}=2\left(2+\sqrt{2}\right)$

$x=\sqrt{2}.\sqrt{2+\sqrt{2}}$

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