Question

Bài 4: Cho biểu thức: D =\(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
a) Rút gọn biểu thức D.
b) Chứng minh răng: 0 <D< 2

Answer 1

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

a) \(D=\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}-1}{2}\)

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

b) Ta có: \(x+\sqrt{x}+1>1\)

Suy ra \(D=\frac{2}{x+\sqrt{x}+1}< 2\)

\(\Rightarrow0< D< 2\)