Question

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

a) tìm điều kiện để A có nghĩa

b) rút gọn A

c) tìm x để A =\(\frac{1}{5}\)

Answer 1

a) ĐKXD : \(x\ge0;x\ne1\)

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

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

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

\(A=\frac{1}{x+1}\)

c) \(A=\frac{1}{5}\Rightarrow\frac{1}{x+1}=\frac{1}{5}\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)