Nếu \(x-2\ge0\Rightarrow x\ge2\Rightarrow\left|x-2\right|=x-2\)
Ta có phương trình
\(x^2-6\left(x-2\right)-4=0\)
\(\Rightarrow x^2-6x+12-4=0\)
\(\Rightarrow x^2-6x+8=0\)
\(\Rightarrow x^2-2x-4x+8=0\)
\(\Rightarrow x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Rightarrow x-2=0\) hoặc \(x-4=0\)
\(\left(+\right)x-2=0\Rightarrow x=2\) (tm)
\(\left(+\right)x-4=0\Rightarrow x=4\) (tm)
Nếu \(x-2<0\Rightarrow x<2\Rightarrow\left|x-2\right|=-\left(x-2\right)=2-x\)
Ta có phương trình
\(x^2-6\left(2-x\right)-4=0\)
\(\Rightarrow x^2-12+6x-4=0\)
\(\Rightarrow x^2+6x-16=0\)
\(\Rightarrow x^2-2x+8x-16=0\)
\(\Rightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+8\right)=0\)
\(\Rightarrow x-2=0\) hoặc \(x+8=0\)
\(\left(+\right)x-2=0\Rightarrow x=2\) (không tm)
\(\left(+\right)x+8=0\Rightarrow x=-8\left(tm\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-8;2;4\right\}\)
