Đáp án:
1)
$\left[\begin{array}{l}x=-\dfrac{5}{3}\\x=\dfrac{11}{5}\end{array}\right.$
2)
$\left[\begin{array}{l}x=3\\x=\dfrac{1}{2}\end{array}\right.$
3)
$\left[\begin{array}{l}x=2\\x=1\\x=-\dfrac{8}{3}\end{array}\right.$
Giải thích các bước giải:
1)
$(3x+5)(2x-7)-(3x+5)(4-3x)=0$
$⇔(3x+5)[2x-7-(4-3x)]=0$
$⇔(3x+5)(2x-7-4+3x)=0$
$⇔(3x+5)(5x-11)=0$
$⇔\left[\begin{array}{l}3x+5=0\\5x-11=0\end{array}\right.$
$⇔\left[\begin{array}{l}x=-\dfrac{5}{3}\\x=\dfrac{11}{5}\end{array}\right.$
2)
$7x(x-3)=x^2-9$
$⇔7x(x-3)-(x-3)(x+3)=0$
$⇔(x-3)[7x-(x+3)]=0$
$⇔(x-3)(6x-3)=0$
$⇔\left[\begin{array}{l}x-3=0\\6x-3=0\end{array}\right.$
$⇔\left[\begin{array}{l}x=3\\x=\dfrac{1}{2}\end{array}\right.$
3)
$x^3-8=(2-x)(2x^2+3x-12)$
$⇔(x-2)(x^2+2x+4)-(2-x)(2x^2+3x-12)=0$
$⇔(x-2)(x^2+2x+4)+(x-2)(2x^2+3x-12)=0$
$⇔(x-2)(x^2+2x+4+2x^2+3x-12)=0$
$⇔(x-2)(3x^2+5x-8)=0$
$⇔(x-2)(x-1)\Bigg(x+\dfrac{8}{3}\Bigg)=0$
$⇔\left[\begin{array}{l}x-2=0\\x-1=0\\x+\dfrac{8}{3}=0\end{array}\right.$
$⇔\left[\begin{array}{l}x=2\\x=1\\x=-\dfrac{8}{3}\end{array}\right.$
