Giải thích các bước giải:
a.Ta có:
$(x+6)(3x-1)+x^2-36=0$
$\to (x+6)(3x-1)+(x+6)(x-6)=0$
$\to (x+6)(3x-1+x-6)=0$
$\to (x+6)(4x-7)=0$
$\to x\in\{-6, \dfrac74\}$
b.Ta có:
$(x+1)(2x-3)(3x+2)=0$
$\to x+1=0\to x=-1$
Hoặc $2x-3=0\to 2x=3\to x=\dfrac32$
Hoặc $3x+2=0\to 3x=-2\to x=-\dfrac23$
$\to x\in\{-1,\dfrac32,-\dfrac23\}$
c.Ta có:
$x^3+2x^2-x-2=0$
$\to x^2(x+2)-(x+2)=0$
$\to (x^2-1)(x+2)=0$
$\to (x-1)(x+1)(x+2)=0$
$\to x-1=0\to x=1$
Hoặc $x+1=0\to x=-1$
Hoặc $x+2=0\to x=-2$
$\to x\in\{1,-1,-2\}$
d.Ta có:
$2x^3-7x^2+7x-2=0$
$\to (2x^3-2)-(7x^2-7x)=0$
$\to 2(x^3-1)-7(x^2-x)=0$
$\to 2(x-1)(x^2+x+1)-7x(x-1)=0$
$\to (x-1)(2x^2+2x+2-7x)=0$
$\to (x-1)(2x^2-5x+2)=0$
$\to (x-1)(2x^2-4x-x+2)=0$
$\to (x-1)(2x-1)(x-2)=0$
$\to x\in\{1,\dfrac12,2\}$
