ĐKXĐ: ...
Đặt \(\sqrt{\frac{4x+9}{28}}=\frac{2t+1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{4x+9}{28}=\frac{\left(2t+1\right)^2}{4}\\\frac{2t+1}{2}=7\left(x^2+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=14t^2+14t\\2t+1=14x^2+14x\end{matrix}\right.\)
Trừ vế cho vế: \(14x^2-14t^2+14x-14t=2t-2x\)
\(\Leftrightarrow14\left(x-t\right)\left(x+t\right)+16\left(x-t\right)=0\)
\(\Leftrightarrow\left(x-t\right)\left(14x+14t+16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=t\\14x+14t+16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{\frac{4x+9}{7}}=2x+1\left(x\ge-\frac{1}{2}\right)\\\sqrt{28x+49}=-14x-9\left(x\le-\frac{9}{14}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4x+9}{7}=\left(2x+1\right)^2\left(x\ge-\frac{1}{2}\right)\\28x+49=\left(-14x-9\right)^2\left(x\le-\frac{9}{14}\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
