Đề trước bị nhầm ạ bạn giải rõ hơn đc ko ?
3x^2+3x(x-2)-6(x^2+3x-1)=0
=>3x^2+3x^2-6x-6x^2-6x+6=0
=>6x+6=0
=>6x=6
=>x=1
\(3x^2+3x\left(x-2\right)-6\left(x^2+3x-1\right)=0\)
\(\Rightarrow\)\(3x^2+3x^2-6x-6x^2-18x+6=0\)
\(\Rightarrow\)\(-24x+6=0\)
\(\Rightarrow\)\(-24x=-6\)
\(\Rightarrow\)\(x=\frac{1}{4}\)
\(3x^2+3x\left(x-2\right)-6\left(x^2+3x-1\right)=0\)
\(3x^2+3x^2-6x-6x^2-18x+6=0\)
\(\left(3x^2+3x^2-6x^2\right)-\left(6x+18\right)+6=0\)
\(-24x+6=0\)
\(-24=-6\)
\(x=\frac{-6}{-24}\)
\(x=\frac{1}{4}\)
