Question

giải chi cho mình với nha:

Answer 1

a: ĐKXĐ: a<>-1

Để \(\dfrac{a^2+a+3}{a+1}\in Z\) thì \(a^2+a+3⋮a+1\)

=>\(a\left(a+1\right)+3⋮a+1\)

=>\(3⋮a+1\)

=>\(a+1\in\left\{1;-1;3;-3\right\}\)

=>\(a\in\left\{0;-2;2;-4\right\}\)

b: x-2xy+y=0

=>\(x\left(1-2y\right)+y-0,5=-0,5\)

=>\(-2x\left(y-0,5\right)+\left(y-0,5\right)=-0,5\)

=>\(\left(y-0,5\right)\left(-2x+1\right)=-0,5\)

=>\(\left(2y-1\right)\left(-2x+1\right)=-1\)

=>(2y-1)(2x-1)=1

=>\(\left(2x-1;2y-1\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(1;1\right);\left(0;0\right)\right\}\)