Question

P=\(\left(\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\):\(\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
a,Rút gọn P
b.Tìm x để \(\frac{1}{P}\) ≤\(-\frac{5}{2}\)

Answer 1

ĐKXĐ: \(x\ge0;x\ne4;9\)

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

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

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

\(\frac{1}{P}\le-\frac{5}{2}\Leftrightarrow\frac{1}{P}+\frac{5}{2}\le0\Leftrightarrow\frac{x-4}{\sqrt{x}+1}+\frac{5}{2}\le0\)

\(\Leftrightarrow\frac{2x+5\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\Leftrightarrow2x+5\sqrt{x}-3\le0\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\le0\Leftrightarrow2\sqrt{x}-1\le0\)

\(\Rightarrow\sqrt{x}\le\frac{1}{2}\Rightarrow0\le x\le\frac{1}{4}\)

Answer 2

tth giúp mk vs...