\(\left(2x-2\right)^2=\left(x+1\right)^2+3.\left(x-2\right)\left(x+5\right)\)
\(\left(2x-2\right)^2-\left(x+1\right)^2=3.\left(x-2\right)\left(x+5\right)\)
\(\left(2x-2-x-1\right)\left(2x-2+x+1\right)=3.\left(x-2\right)\left(x+5\right)\)
\(\left(x-3\right)\left(3x-1\right)=3.\left(x-2\right)\left(x+5\right)\)
\(3x^2-x-9x+3=\left(3x-6\right)\left(x+5\right)\)
\(3x^2-10x+3=3x^2+15x-6x-30\)
\(3x^2-3x^2-10x+6x-15x+3+30=0\)
\(-19x+33=0\)
\(-19x=-33\)
\(x=\frac{33}{19}\)
\(\Leftrightarrow\left(2x-2\right)^2-\left(x+1\right)^2-3.\left(x-2\right).\left(x+5\right)=0\)
\(\Leftrightarrow4x^2-8x+4-\left(x^2+2x+1\right)-\left(3x-6\right).\left(x+5\right)=0\)
\(\Leftrightarrow4x^2-8x+4-x^2-2x-1-\left(3x^2+15x-6x-30\right)=0\)
\(\Leftrightarrow4x^2-8x+4-x^2-2x-1-3x^2-15x+6x+30=0\)
\(\Leftrightarrow-19x+33=0\)
\(\Leftrightarrow-19x=-33\)
\(\Leftrightarrow x=\frac{33}{19}\)
Vậy...............
