\(\left(x+2\right)^2-3\left(x-3\right)=\left(x-1\right)^2\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-1\right)^2-3\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+2+x-1\right)\left(x+2-x+1\right)-3\left(x-3\right)=0\)
\(\Leftrightarrow3\left(2x-1\right)-3\left(x-3\right)=0\)
\(\Leftrightarrow3\left(2x-1-x+3\right)=0\)
\(\Leftrightarrow3\left(x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy...
(x+2)2 - 3(x-3) = (x-1)2
⇔ x2 + 4x + 4 - 3x + 9 = x2 - 2x + 1
⇔ x2 + 4x + 4 - 3x+ 9 - x2 + 2x -1 =0
⇔ 3x + 12 = 0
⇔ 3(x+4) = 0
⇔ x + 4 = 0
⇔ x = -4
