1: Ta có: \(x^2+2x-4=-12+3x+x^2\)
\(\Leftrightarrow x^2+2x-4+12-3x-x^2=0\)
\(\Leftrightarrow-x+8=0\)
\(\Leftrightarrow-x=-8\)
hay x=8
Vậy: x=8
2: Ta có: \(x\left(2x-3\right)-x^2+2=x\left(x-5\right)-1\)
\(\Leftrightarrow2x^2-3x-x^2+2=x^2-5x-1\)
\(\Leftrightarrow x^2-3x+2-x^2+5x+1=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
hay \(x=\frac{-3}{2}\)
Vậy: \(x=\frac{-3}{2}\)
3: Ta có: \(\left(x-1\right)\left(x+3\right)=x^2-4\)
\(\Leftrightarrow x^2+3x-x-3-x^2+4=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
hay \(x=\frac{-1}{2}\)
Vậy: \(x=\frac{-1}{2}\)
4: Ta có: \(\left(6x+2\right)\left(x-2\right)=2x\left(3x-5\right)\)
\(\Leftrightarrow6x^2-12x+2x-4=6x^2-10x\)
\(\Leftrightarrow6x^2-10x-4-6x^2+10x=0\)
\(\Leftrightarrow-4< 0\)
Vậy: x∈∅
5: Ta có: \(\left(x-2\right)^2=\left(x-3\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-4x+4=x^2+2x-3x-6\)
\(\Leftrightarrow x^2-4x+4-x^2-2x+3x+6=0\)
\(\Leftrightarrow-3x+10=0\)
\(\Leftrightarrow-3x=-10\)
hay \(x=\frac{10}{3}\)
Vậy: \(x=\frac{10}{3}\)
