help me
Câu 2:
a: \(\Leftrightarrow4x^2+4x+1-4x^2-3x=2x-2\)
=>x+1=2x-2
=>-x=-3
hay x=3
b: \(\Leftrightarrow x\left(x^2-2x-24\right)=0\)
\(\Leftrightarrow x\left(x-6\right)\left(x+4\right)=0\)
hay \(x\in\left\{0;6;-4\right\}\)
c: \(\Leftrightarrow x^2-3x+2-3x-6=-x-10\)
\(\Leftrightarrow-6x-4=-x-10\)
=>-5x=-6
hay x=6/5(nhận)
Câu 2 :
a, \(4x^2+4x+4-4x^2-3x=2x-2\Leftrightarrow x+4=2x-2\Leftrightarrow x=6\)
b, \(x\left(x^2-2x+1-25\right)=0\Leftrightarrow x\left(x-6\right)\left(x+4\right)=0\Leftrightarrow x=0;x=6;x=-4\)
c, đk : x khác 2 ; -2 \(\Rightarrow x^2-3x+2-3x-6=-x-10\)
\(\Leftrightarrow x^2-6x-4=-x-10\Leftrightarrow x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow x=2\left(ktm\right);x=3\)