Câu 1 :
a) $2(x-3)-3(x-5)=4(3-x)-18$
$⇔2x-6-3x+15=12-4x-18$
$⇔-x+9=-6-4x$
$⇔3x=-15$
$⇔x=-5$
b) $-2x-11 \vdots 3x+2$
⇔$⇔6x+33 \vdots 3x+2$
$⇔2.(3x+2)+29 \vdots 3x+2$
$⇔29 \vdots 3x+2$
$⇔3x+2 ∈ Ư(29)=${$-1,1,-29,29$}
$⇔x∈ ${$-1,9, \frac{1}{3},\frac{-31}{3}$}
Câu 2 :
$-a(c-d)-d(a+c)=-ac+ad-ad-dc=-ac-dc=-c(a+d)$
Câu 3 :
$(3a+2)(2a-1)+(3-a)(6a+2)-17(a-1)$
$=6a^2+a-2+18a+6-6a^2-2a-17a+17$
$=21$ không phụ thuộc vào a
Câu 4 :
Để :$a^2=2a$
$⇔a=0,a=2$
Để $a^2>2a$
$⇔a>2$
Để $a^2<2a$
$⇔a<2$