Các bạn k cho mình đi rùi mình
Giải cho
a) \(ĐKXĐ:\hept{\begin{cases}x\ne9\\x\ge0\end{cases}}\)
\(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow C=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow C=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}-3}\)
\(\Leftrightarrow C=\frac{3}{3-\sqrt{x}}\cdot\frac{\sqrt{x}-3}{2\sqrt{x}+4}\)
\(\Leftrightarrow C=\frac{-3}{2\sqrt{x}+4}\)
b) Để \(C< -1\)
\(\Leftrightarrow\frac{-3}{2\sqrt{x}+4}< -1\)
\(\Leftrightarrow\frac{-3+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}+1}{2\sqrt{x}+4}< 0\) (1)
Vì \(\hept{\begin{cases}2\sqrt{x}+1>0\\2\sqrt{x}+4>0\end{cases}}\)
\(\Leftrightarrow\frac{2\sqrt{x}+1}{2\sqrt{x}+4}>0\) (2)
Từ (1) và (2) suy ra : Để \(C< -1\Leftrightarrow x\in\varnothing\)