Bài 2:
a, \(3\left(x-1\right)^2-\left(x+2\right)^2=2x^2-5\)
\(\Leftrightarrow3\left(x^2-2x+1\right)-\left(x^2+4x=4\right)=2x^2-5\)
\(\Leftrightarrow3x^2-6x+3-x^2-4x-4=2x^2-5\)
\(\Leftrightarrow2x^2-10x-1=2x^2-5\)
\(\Leftrightarrow-10x-1=-5\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
b,\(\left(2x+3\right)^2-\left(x-3\right)^2=3x\left(x+2\right)\)
\(\Leftrightarrow4x^2+12x=9-\left(x^2-6x=9\right)=3x^2+6x\)
\(\Leftrightarrow4x^2+12x+9-x^2+6x-9=3x^2+6x\)
\(\Leftrightarrow4x^2+12x-x^2=3x^2\)
\(\Leftrightarrow3x^2+12x=3x^2\)
\(\Leftrightarrow12x=0\)
\(\Leftrightarrow x=0\)
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