Question

Cho biểu thức:

A=\(\left(\frac{2\sqrt{x}}{\sqrt{x^3}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(2+\frac{2\sqrt{x}}{x+1}\right)\), với \(x\ge0\); \(x\ne1\)

a) Rút gọn biểu thức A

b) Tìm giá trị của x để \(A\le0\)

Answer 1

ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

\(A\le0\Leftrightarrow\frac{1-\sqrt{x}}{2\left(x+\sqrt{x}+1\right)}\le0\)

\(\Leftrightarrow1-\sqrt{x}\le0\) (do \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\))

\(\Leftrightarrow x\ge1\)

Kết hợp ĐKXĐ ta được \(x>1\)