Question

Cho biểu thức Q = \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\), với x > 0, x # 1

a, Rút gọn biểu thức Q

Answer 1

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

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

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

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

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

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

Q = \(\dfrac{2\sqrt{x}}{x-1}\sqrt{x}\) = \(\dfrac{2\sqrt{x}+x\sqrt{x}-\sqrt{x}}{x-1}\) = \(\dfrac{x\sqrt{x}+\sqrt{x}}{x-1}\)

Answer 2

Hữu tỷ hóa bằng cách đặt \(t=\sqrt{x}\) thì

\(Q=\left(\dfrac{t+2}{\left(t+1\right)^2}-\dfrac{t-2}{t^2-1}\right)\left(t^2+t\right)=\dfrac{\left(t+2\right)\left(t-1\right)-\left(t-2\right)\left(t+1\right)}{\left(t+1\right)^2\left(t-1\right)}t\left(t+1\right)\)

\(=\dfrac{2t^2}{\left(t+1\right)\left(t-1\right)}=\dfrac{2t^2}{t^2-1}=\dfrac{2x}{x-1}\)