a.
ĐKXĐ: \(x\ne-3\)
\(\Leftrightarrow x^2+\dfrac{9x^2}{\left(x+3\right)^2}-\dfrac{6x^2}{x+3}+\dfrac{6x^2}{x+3}=40\)
\(\Leftrightarrow\left(x-\dfrac{3x}{x+3}\right)^2+\dfrac{6x^2}{x+3}-40=0\)
\(\Leftrightarrow\left(\dfrac{x^2}{x+3}\right)^2+\dfrac{6x^2}{x+3}-40=0\)
Đặt \(\dfrac{x^2}{x+3}=t\)
\(\Rightarrow t^2+6t-40=0\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2}{x+3}=4\\\dfrac{x^2}{x+3}=-10\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-12=0\\x^2+10x+30=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
b.
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(\dfrac{4}{4x+\dfrac{7}{x}-8}+\dfrac{3}{4x+\dfrac{7}{x}-10}=1\)
Đặt \(4x+\dfrac{7}{x}-10=t\)
\(\Rightarrow\dfrac{4}{t+2}+\dfrac{3}{t}=1\)
\(\Leftrightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)
\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\dfrac{7}{x}-10=-1\\4x+\dfrac{7}{x}-10=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\left(vô-nghiệm\right)\\4x^2-16x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
