Question

BT: Rút gọn M =\(\left(\frac{2x}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)

                   

Đk: \(x\ge0;x\ne1\)Hộ mik vs mai cần ròi 😫😫

Answer 1

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

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

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

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

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

\(M=\frac{2x}{x^3-1}\)

vậy \(M=\frac{2x}{x^3-1}\)