Question

21.A=\(\left(\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\frac{\sqrt{x}}{\sqrt{x}-1}\right)\)

Answer 1

\(A=\left(\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\frac{\sqrt{x}}{\sqrt{x}-1}\right)\)

\(=\frac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{x-1}:\frac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}\)

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

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

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

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