\(ĐK:x\ne-3\)
\(PT\Leftrightarrow\left[x^2-\dfrac{6x^2}{x+3}+\dfrac{9x^2}{\left(x+3\right)^2}\right]+\dfrac{6x^2}{x+3}-40=0\)
\(\Leftrightarrow\left(x-\dfrac{3x}{x+3}\right)^2+\dfrac{6x^2}{x+3}-40=0\)
\(\Leftrightarrow\left(\dfrac{x^2}{x+3}\right)^2+6.\dfrac{x^2}{x+3}-40=0\)
Đặt \(\dfrac{x^2}{x+3}=a\) \(\Rightarrow PT\Leftrightarrow a^2+6a-40=0\Leftrightarrow\left(a+10\right)\left(a-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{x+3}=-10\\\dfrac{x^2}{x+3}=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+10x+30=0\\x^2-4x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2+5=0\left(l\right)\\\left(x+2\right)\left(x-6\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\) (TMĐKXĐ)
Vậy \(x=-2;x=6\)
\(x^2+\dfrac{9x^2}{\left(x+3\right)^2}=40\)
\(\Leftrightarrow x^2+\dfrac{9x^2}{x^2+6x+9}-40=0\)
\(\Leftrightarrow x^2+9+\dfrac{3}{2}x+x^2-40=0\)
\(\Leftrightarrow2x^2+\dfrac{3}{2}x-31=0\)
\(\Leftrightarrow\dfrac{4x^2}{2}+\dfrac{3x}{2}-\dfrac{62}{2}=0\)
\(\Rightarrow4x^2+3x-62=0\)
\(\Leftrightarrow4\left(x^2+\dfrac{3}{4}x+\dfrac{9}{64}\right)-\dfrac{1001}{16}=0\)
\(\Leftrightarrow4\left(x+\dfrac{3}{8}\right)^2=\dfrac{1001}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{8}=\sqrt{\dfrac{1001}{16}}\\x+\dfrac{3}{8}=-\sqrt{\dfrac{1001}{16}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{1001}{16}}-\dfrac{3}{8}\\x=-\sqrt{\dfrac{1001}{16}}-\dfrac{3}{8}\end{matrix}\right.\)
