Question

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

a) Tìm điều kiện của x để A có nghĩa .

b) Rút gọn A .

Answer 1

\(a,ĐKXĐ:\hept{\begin{cases}x^2-\sqrt{x}\ne0\\x\ge0\\\sqrt{x}+1\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x>0\end{cases}}\)

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

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

\(=\frac{1}{\sqrt{x}\left(\sqrt{x}^3-1\right)}\cdot\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)

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

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