Đáp án:
$\begin{array}{l}(3x-2)^2+2(3x-2)(x+6)+(x-6)^2=0\\\leftrightarrow 9x^2-12x+4+(6x-4)(x+6)+x^2-12x+36=0\\\leftrightarrow 9x^2-12x+4+6x^2+36x-4x-24+x^2-12x+36=0\\\leftrightarrow 16x^2+8x+16=0\\\leftrightarrow 2x^2+x+2=0\\\leftrightarrow 2x^2+x=-2\\\leftrightarrow x^2+\dfrac{1}{2}x=-1\\\leftrightarrow x^2+\dfrac{1}{2}x+\dfrac{1}{16}=-1+\dfrac{1}{16}\\\leftrightarrow \left(x+\dfrac{1}{4}\right)^2=-\dfrac{15}{16} \ (\text{vô nghiệm,} \ \left(x+\dfrac{1}{4}\right)^2\geqslant0\forall x)\\\leftrightarrow x\in\varnothing\\\text{Vậy} \ S=\varnothing\\\text{Khác:}\\(3x-2)^2+2(3x-2)(x-6)+(x-6)^2=0\\\leftrightarrow (3x-2+x-6)^2=0\\\leftrightarrow (4x-8)^2=0\\\leftrightarrow 4x-8=0\\\leftrightarrow x=2\\\text{Vậy} \ S=\{2\}\end{array}$
