1: Ta có: \(\left(6x+2\right)\left(x-2\right)=2x\left(3x-5\right)\)
\(\Leftrightarrow6x^2-12x+2x-4=6x^2-10x\)
\(\Leftrightarrow6x^2-10x-4-6x^2+10x=0\)
\(\Leftrightarrow-4\ne0\)
Vậy: Tập nghiệm S=∅
2: Ta có: \(\left(x-2\right)^2=\left(x-3\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-4x+4=x^2+2x-3x-6\)
\(\Leftrightarrow x^2-4x+4=x^2-x-6\)
\(\Leftrightarrow x^2-4x+4-x^2+x+6=0\)
\(\Leftrightarrow-3x+10=0\)
\(\Leftrightarrow-3x=-10\)
hay \(x=\frac{10}{3}\)
Vậy: \(x=\frac{10}{3}\)
Đúng 0
Bình luận (0)