Question

Give this problem a try and try to solve this​

Answer 1

Answer:

No solution

Step-by-step explanation:

Given equation is,

$\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1+x^\frac{1}{2}}-\frac{(4+x)^\frac{1}{2}}{(1-x)^\frac{1}{2}}=0$

$\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1+x^\frac{1}{2}}=\frac{(4+x)^\frac{1}{2}}{(1-x)^\frac{1}{2}}$

$\frac{(x+1)}{\sqrt{x}(1-x)}+\frac{(\sqrt{x}-1)}{\sqrt{x}(1+\sqrt{x})}=(\frac{4+x}{1-x})^{\frac{1}{2}}$

$\frac{(\sqrt{x}+1)(x+1)+(\sqrt{x}-1)(1-x)}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^{\frac{1}{2}}$

$\frac{x\sqrt{x}+x+\sqrt{x}+1+\sqrt{x}-1-x\sqrt{x}+x}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}$

$\frac{2x+2\sqrt{x}}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}$

$\frac{2(\sqrt{x}+1)}{(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}$

$\frac{2}{1-x}=(\frac{4+x}{1-x})^\frac{1}{2}$  if x ≠ ±1

$(\frac{2}{1-x})^2=\frac{4+x}{1-x}$  [Squaring on both the sides of the equation]

$\frac{4}{(1-x)}=(4+x)$

4 = (1 - x)(4 + x)

4 = 4 - 4x + x - x²

0 = -3x - x²

x² + 3x = 0

x(x + 3) = 0

x = 0, -3

But both the solutions x = 0 and x = -3 are extraneous solutions, given equation has no solution.

Answer 2

Answer:

Could you please help me Genius??????