Question

cho biểu thức A= \(\frac{1}{x^2-\sqrt{x}}:\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\)

rút gọn A

Answer 1

Ta có: \(A=\frac{1}{x^2-\sqrt{x}}:\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\)

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

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

\(=\frac{1}{x-1}\)

Answer 2

ĐKXĐ: ...

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

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

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