Đáp án:a.$(x-1)(x+4)(x^2+3x+6)$
b.$x\in\{-\dfrac{10}{3},1\}$
Giải thích các bước giải:
Ta có:
$(x^2+2x)(x^2+4x+3)-24$
$=x(x+2)(x^2+3x+x+3)-24$
$=x(x+2)(x(x+3)+x+3)-24$
$=x(x+2)(x+1)(x+3)-24$
$=(x(x+3))((x+2)(x+1))-24$
$=(x^2+3x)(x^2+3x+2)-24$
$=(x^2+3x)^2+2(x^2+3x)-24$
$=(x^2+3x)^2+2(x^2+3x)+1-25$
$=(x^2+3x+1)^2-5^2$
$=(x^2+3x+1-5)(x^2+3x+1+5)$
$=(x^2+3x-4)(x^2+3x+6)$
$=(x^2+4x-x-4)(x^2+3x+6)$
$=(x-1)(x+4)(x^2+3x+6)$
b.Ta có:
$3x^2+7x=10$
$\to 3x^2+7x-10=0$
$\to 3x^2-3x+10x-10=0$
$\to 3x(x-1)+10(x-1)=0$
$\to (3x+10)(x-1)=0$
$\to x\in\{-\dfrac{10}{3},1\}$
