B1
2x+12=3(x-7)
=>2x+12=3x-21
=>x=33
Vậy x=33
a)A=(-a-b+c)-(a-b-c)=-a-b+c-a+b+c=-2a+2c+0b
b, Thay a=1;b=-1;c=-2 vào A ta có:-2*1+2*1+0*-1=0
Vậy A = 0 khi a=1;b=-1;c=-2
B2
Ta có:$\frac{6a+1}{3a-1}$ =$\frac{6a-2+3}{3a-1}$ =$\frac{2(3a-1)+3}{3a-1}$ =2+ $\frac{3}{3a-1}$ Vì a∈Z nên 3 chia hết cho 3a-1 => 3a-1 ∈ Ư(3)={±1;±2}
=>a∈{0;1}
Vậy a∈{0;1}