Sửa đề: \(G=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2-2\sqrt{x}}{3-\sqrt{x}}-1\right)\)
a) Ta có: \(G=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2-2\sqrt{x}}{3-\sqrt{x}}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)
\(=\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3}{\sqrt{x}+3}\)
b) Để \(G< \frac{1}{3}\) thì \(G-\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{-3}{\sqrt{x}+3}-\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{-9}{3\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\frac{-9-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow-12-\sqrt{x}< 0\)(Vì \(3\left(\sqrt{x}+3\right)>0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\sqrt{x}+12>0\)(Luôn đúng với mọi x thỏa mãn ĐKXĐ)
