Đáp án:
Giải thích các bước giải:
$a)x.(x-2)-x^2+3x=4$
$⇔x^2-2x-x^2+3x=4$
$⇔x=4$
$b)3x^2-3x=(x-1)^2$
$⇔3x^2-3x=x^2-2x+1$
$⇔3x^2-3x-x^2+2x-1=0$
$⇔2x^2-x-1=0$
$⇔(2x^2-2x)+(x-1)=0$
$⇔2x.(x-1)+(x-1)=0$
$⇔(x-1).(2x+1)=0$
$⇔$\(\left[ \begin{array}{l}x-1=0\\2x+1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{2}\end{array} \right.\)
$c)(x+2).(x^2-2x+4)-x.(x-2)^2=-12$
$⇔x^3+8-x.(x^2-4x+4)+12=0$
$⇔x^3+8-x^3+4x^2-4x+12=0$
$⇔4x^2-4x+20=0$
$⇔(2x)^2-2.2x.1+1+19=0$
$⇔(2x-1)^2+19=0 (vô lí)$
⇒ PT vô nghiệm
