Question

Answer 1

a) a²+ab+8a+8b
= (a²+8a)+(ab+8b)
= a(a+8)+b(a+8)
= (a+b)(a+8)
b) 5x(x+8)-3(x+8)=0
(5x-3)(x+8)=0
Th1: 5x-3=0                Th2: x+8=0
        5x=3                                x=-8
          x=3/5
Vậy x\(\in\)\(\left\{\dfrac{3}{5};-8\right\}\)
c) \(\dfrac{1}{ah-a^2}-\dfrac{1}{h^2-ah}=\dfrac{1}{a\left(h-a\right)}-\dfrac{1}{h\left(h-a\right)}=\dfrac{h}{ah\left(h-a\right)}-\dfrac{a}{ah\left(h-a\right)}=\dfrac{h-a}{ah\left(h-a\right)}=\dfrac{1}{ah}\)