Solve for x 1/( square root of 2+x- square root of 2-x)+1/( square root of 2+x+ square root of 2-x)=1

$\frac{1}{\sqrt{2 + x} - \sqrt{2 - x}} + \frac{1}{\sqrt{2 + x} + \sqrt{2 - x}} = 1$

Simplify each term.

Multiply $\frac{1}{\sqrt{2 + x} - \sqrt{2 - x}}$ by $\frac{\sqrt{2 + x} + \sqrt{2 - x}}{\sqrt{2 + x} + \sqrt{2 - x}}$ .

$\frac{1}{\sqrt{2 + x} - \sqrt{2 - x}} \cdot \frac{\sqrt{2 + x} + \sqrt{2 - x}}{\sqrt{2 + x} + \sqrt{2 - x}} + \frac{1}{\sqrt{2 + x} + \sqrt{2 - x}} = 1$

Multiply $\frac{1}{\sqrt{2 + x} - \sqrt{2 - x}}$ by $\frac{\sqrt{2 + x} + \sqrt{2 - x}}{\sqrt{2 + x} + \sqrt{2 - x}}$ .

$\frac{\sqrt{2 + x} + \sqrt{2 - x}}{(\sqrt{2 + x} - \sqrt{2 - x}) (\sqrt{2 + x} + \sqrt{2 - x})} + \frac{1}{\sqrt{2 + x} + \sqrt{2 - x}} = 1$

Expand the denominator using the FOIL method.

$\frac{\sqrt{2 + x} + \sqrt{2 - x}}{{\sqrt{2 + x}}^{2} + \sqrt{(2 + x) (2 - x)} - \sqrt{(2 + x) (2 - x)} - {\sqrt{2 - x}}^{2}} + \frac{1}{\sqrt{2 + x} + \sqrt{2 - x}} = 1$

Simplify.

$\frac{\sqrt{2 + x} + \sqrt{2 - x}}{2 x} + \frac{1}{\sqrt{2 + x} + \sqrt{2 - x}} = 1$

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x = 2$

Solve for x 1/( square root of 2+x- square root of 2-x)+1/( square root of 2+x+ square root of 2-x)=1

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