\(6\left(x-\frac{x}{x+1}\right)^2+\frac{x^2-12x-12}{x+1}=0\) (*)
ĐKXĐ: \(x+1\ne0\Leftrightarrow x\ne-1\)
Ta có: \(\left(x-\frac{x}{x+1}\right)^2\ge0\forall x\Rightarrow6\left(x-\frac{x}{x+1}\right)^2\ge0\forall x\)
\(\Rightarrow6\left(x-\frac{x}{x+1}\right)^2+\frac{x^2-12x-12}{x+1}\ge\frac{x^2-12x-12}{x+1}\forall x\)
Dấu "=" xảy ra khi: \(\frac{x^2-12x-12}{x+1}=0\)
\(\Leftrightarrow x^2-12x-12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6+4\sqrt{3}\\x=6-4\sqrt{3}\end{matrix}\right.\) (t/m ĐKXĐ)
Vậy phương trình có tập nghiệm là S = \(\left\{6+4\sqrt{3};6-4\sqrt{3}\right\}\)
\(6\cdot\left(x-\frac{x}{x+1}\right)^2+\frac{x^2-12x-12}{x+1}=0\)
\(\Leftrightarrow6x^3+7x^2-\frac{12x^3}{x+1}-\frac{12x^2}{x+1}+\frac{6x^3}{\left(x+1\right)^2}+\frac{6x^2}{\left(x+1\right)^2}-12x-12=0\)
\(\Leftrightarrow6x^3\cdot\left(x+1\right)^2+7x^2\cdot\left(x+1\right)^2-12x^3\cdot\left(x+1\right)-12x^2\cdot\left(x+1\right)+6x^3+6x^2-12x\cdot\left(x+1\right)^2-12\cdot\left(x+1\right)^2=0\)
\(\Leftrightarrow6x^5+7x^4-10x^3-35x^2-36x-12=0\)
\(\Leftrightarrow\left(6x^4+x^3-11x^2-24x-12\right)\cdot\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x^3+13x^2+15x+6\right)\cdot\left(x-2\right)\cdot\left(x+1\right)=0\)
\(\Rightarrow x=\left[{}\begin{matrix}-1\\2\\-\frac{2}{3}\end{matrix}\right.\)
