Answer:
No solution
Step-by-step explanation:
Given equation is,
[tex]\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[/tex]
[tex]\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}}[/tex]
[tex]\frac{(x+1)}{\sqrt{x}(1-x)}+\frac{(\sqrt{x}-1)}{\sqrt{x}(1+\sqrt{x})}=(\frac{4+x}{1-x})^{\frac{1}{2}}[/tex]
[tex]\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}}[/tex]
[tex]\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}[/tex]
[tex]\frac{2x+2\sqrt{x}}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}[/tex]
[tex]\frac{2(\sqrt{x}+1)}{(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}[/tex]
[tex]\frac{2}{1-x}=(\frac{4+x}{1-x})^\frac{1}{2}[/tex] if x ≠ ±1
[tex](\frac{2}{1-x})^2=\frac{4+x}{1-x}[/tex] [Squaring on both the sides of the equation]
[tex]\frac{4}{(1-x)}=(4+x)[/tex]
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.
