Question

rút gọn biểu thức: P=\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\): \(\dfrac{2\sqrt{x}+1}{x-\sqrt{x}}\)

Answer 1

\(P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{2\sqrt{x}+1}{x-\sqrt{x}}\left(dkxd:x>0;x\ne1\right)\)

\(=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)

\(=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)

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