Bài 1:
-10-(x-5)+(3-x)=-8
⇒-10-x+5+3-x=-8
⇒-2-2x=-8
⇒2x=6
⇒x=3
Bài 2:
Ta có: -3n+2$\vdots$2n+1
⇒-2(-3n+2)$\vdots$2n+1
⇒6n-4$\vdots$2n+1
⇒3(2n+1)-7$\vdots$2n+1
⇒2n+1∈Ư(7)={±1;±7}
2n+1 1 -1 7 -7
2n 0 -2 6 -8
n 0 1 3 -4
Vậy n∈{0;1;3;-4}
Bài 3:
(3-x)(xy+5)=-1
TH1: $\left \{ {{3-x=1} \atop {xy+5=-1}} \right.$ =>$\left \{ {{x=2} \atop {y=-3}} \right.$
TH2: $\left \{ {{3-x=-1} \atop {xy+5=1}} \right.$ =>$\left \{ {{x=4} \atop {y=-1}} \right.$
Vậy (x,y)∈{(2;-3);(4;-1)}