a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne\frac{9}{4}\end{matrix}\right.\)
Ta có: \(Q=\frac{\sqrt{x}+2}{-\sqrt{x}+2}+\frac{3\sqrt{x}-4}{2\sqrt{x}-3}+\frac{-7\sqrt{x}+10}{-2x+7\sqrt{x}-6}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(2\sqrt{x}-3\right)}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}+\frac{\left(3\sqrt{x}-4\right)\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}+\frac{-7\sqrt{x}+10}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{2x+\sqrt{x}-6-3x+10\sqrt{x}-8-7\sqrt{x}+10}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{-x+4\sqrt{x}-4}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{-\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}-2}{2\sqrt{x}-3}\)
b) Để Q<-4 thì Q+4<0
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}-3}+\frac{4\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
Trường hợp 1: \(\left\{{}\begin{matrix}9\sqrt{x}-14>0\\2\sqrt{x}-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}>14\\2\sqrt{x}< 3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>\frac{14}{9}\\\sqrt{x}< \frac{3}{2}\end{matrix}\right.\)
⇔Loại vì \(\frac{14}{9}>\frac{3}{2}\)
Trường hợp 2: \(\left\{{}\begin{matrix}9\sqrt{x}-14< 0\\2\sqrt{x}-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}< 14\\2\sqrt{x}>3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}< \frac{14}{9}\\\sqrt{x}>\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< \frac{196}{81}\\x>\frac{9}{4}\end{matrix}\right.\Leftrightarrow\frac{9}{4}< x< \frac{196}{81}\)
Kết hợp ĐKXĐ, ta được:
\(\frac{9}{4}< x< \frac{196}{81}\)
Vậy: Để Q<-4 thì \(\frac{9}{4}< x< \frac{196}{81}\)
