a,A=x^2+5x+x+5/x^2+5x-2x+10
=(x+1)(x+5)/(x+2)(x+5)
=(x+1)/(x+2)
b, để A=0 => x+1=0
=>x= -1
a: ĐKXĐ: x<>-5; x<>2
\(A=\dfrac{\left(x+1\right)\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{x+1}{x-2}\)
b: Để A=0 thì x+1=0
=>x=-1
c: Để A là số nguyên thì \(x-2+3⋮x-2\)
=>\(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;-1\right\}\)