Question

P=(\(1-\frac{\sqrt{x}}{\sqrt{x}+1}\)) : \(\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

a) rút gọn P

b) tính giá trị của a để P <0

Answer 1

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

\(P=\frac{1}{\sqrt{x}+1}:\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(P=\frac{1}{\left(\sqrt{x}+1\right)}:\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(P=\frac{1}{\left(\sqrt{x}+1\right)}:\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)

Để \(P< 0\Rightarrow\sqrt{x}-4< 0\Rightarrow x< 4\Rightarrow0\le x< 4\)