Bài 7:
\(2n-3⋮n+1\)
=>\(2n+2-5⋮n+1\)
=>\(n+1\in\left\{1;-1;5;-5\right\}\)
=>\(n\in\left\{0;-2;4;-6\right\}\)
mà n>0
nên n=4
Bài 8:
x+y-2=0
=>x+y=2
\(M=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)
\(=x^2\left(x+y\right)-y\left(x+y\right)-2x^2+2y+x+y+2017\)
\(=2x^2-2y-2x^2+2y+2+2017\)
=2019
