Question

Rút gọn A=(\(\dfrac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}:\dfrac{\sqrt{x}-1}{x+1}\)

Answer 1

\(A=\left(\dfrac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{x+1}\left(x\ge0;x\ne\pm1\right)\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{x+1}{\sqrt{x}-1}\)

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

#\(Toru\)

Answer 2

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$

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

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