Question

bài 1: Cho M= \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

Rút gọn

Answer 1

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

Answer 2

ĐK:x>1

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

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

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

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

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

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