ĐKXĐ: x<>3
\(x^2+\dfrac{9x^2}{\left(x-3\right)^2}=40\)
=>\(\dfrac{x^2\left(x-3\right)^2+9x^2}{\left(x-3\right)^2}=40\)
=>\(x^2\left(x^2-6x+9+9\right)=40\left(x-3\right)^2\)
=>\(x^4-6x^3+18x^2=40\left(x^2-6x+9\right)\)
=>\(x^4-6x^3+18x^2-40x^2+240x-360=0\)
=>\(x^4-6x^3-22x^2+240x-360=0\)
=>\(\left(x-2\right)\left(x+6\right)\left(x^2-10x+30\right)=0\)
mà \(x^2-10x+30=x^2-10x+25+5=\left(x-5\right)^2+5>=5>0\forall x\)
nên (x-2)(x+6)=0
=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
ĐKXĐ: \(x\ne3\)
\(\Leftrightarrow x^2+\left(\dfrac{3x}{x-3}\right)^2+2x.\dfrac{3x}{x-3}-2x.\dfrac{3x}{x-3}=40\)
\(\Leftrightarrow\left(x+\dfrac{3x}{x-3}\right)^2-2x.\dfrac{3x}{x-3}=40\)
\(\Leftrightarrow\left(\dfrac{x^2}{x-3}\right)^2-6.\dfrac{x^2}{x-3}=40\)
Đặt \(\dfrac{x^2}{x-3}=t\)
\(\Rightarrow t^2-6t-40=0\Rightarrow\left[{}\begin{matrix}t=10\\t=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{x-3}=10\\\dfrac{x^2}{x-3}=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=10x-30\\x^2=-4x+12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-10x+30=0\\x^2+4x-12=0\end{matrix}\right.\) (bấm máy)
