Giải thích các bước giải:
Ta có :
(1)$(1-x)(x+2)=0\to x\in\{1,-2\}$
(2)$(2x-2)(6+3x)(3x+2)=0\to 2(x-2)(3(2+x))(3x+2)\to x\in\{1,-2,-\dfrac23\}$
(3)$(5x-5)(3x+2)(8x+4)(x^2-5)=0$
$\to 5x-5=0\to x=1$
$3x+2=0\to x=-\dfrac23$
$8x+4=0\to x=-\dfrac12$
$x^2-5=0\to x=\pm\sqrt{5}$
$\to x\in\{1,-\dfrac23,-\dfrac12,\pm\sqrt5\}$
a.(1)
b.(1)
c.(1),(2)
d.(1),(2),(3)