\(2x^3+3x^2-32x-48=0\)
\(\Leftrightarrow2x^3-32x+3x^2-48=0\)
\(\Leftrightarrow2x\left(x^2-16\right)+3\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm4\\x=-\frac{3}{2}\end{matrix}\right.\)
b/ \(\Leftrightarrow10x^2-15x+4x-6=0\)
\(\Leftrightarrow5x\left(2x^2-3\right)+2\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{2}{5}\end{matrix}\right.\)
Lời giải:
a)
$10x^2-11x-6=0$
$\Leftrightarrow 10x^2-15x+4x-6=0$
$\Leftrightarrow 5x(2x-3)+2(2x-3)=0$
$\Leftrightarrow (2x-3)(5x+2)=0$
$\Rightarrow 2x-3=0$ hoặc $5x+2=0$
$\Rightarrow x=\frac{3}{2}$ hoặc $x=-\frac{2}{5}$
b)
$2x^3+3x^2-32x=48$
$\Leftrightarrow 2x^3+3x^2-32x-48=0$
$\Leftrightarrow 2x^3-8x^2+11x-44x+12x-48=0$
$\Leftrightarrow 2x^2(x-4)+11x(x-4)+12(x-4)=0$
$\Leftrightarrow (x-4)(2x^2+11x+12)=0$
$\Leftrightarrow (x-4)[2x(x+4)+3(x+4)]=0$
$\Leftrightarrow (x-4)(x+4)(2x+3)=0$
$\Rightarrow x-4=0; x+4=0$ hoặc $2x+3=0$
$\Rightarrow x=\pm 4$ hoặc $x=\frac{-3}{2}$
