Question

Cho biểu thức P=\(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\)

a Rút gọn biểu thức P

b Tìm x để P=\(\frac{7}{12}\)

c Tìm x để P>\(\frac{1}{2}\)

Answer 1

ĐKXĐ: \(x\ge0;x\ne4\)

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

Để \(P=\frac{7}{12}\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+3}=\frac{7}{12}\Leftrightarrow12\sqrt{x}-24=7\sqrt{x}+21\)

\(\Leftrightarrow5\sqrt{x}=45\Rightarrow\sqrt{x}=9\Rightarrow x=81\)

Để \(P>\frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+3}>\frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{1}{2}>0\Leftrightarrow\frac{\sqrt{x}-7}{\sqrt{x}+3}>0\)

\(\Leftrightarrow\sqrt{x}-7>0\Rightarrow x>49\)