a/ Thay `x=2` (tm `x`x \neq -5;x\neq-1`) vào bt `B` có
`B=-10/-2=5`
b/ `A=(x+2)/(x+5)+(-5x-1)/(x^2+6x+5)-1/(1+x)` (`x \neq -5;x\neq-1`)
`=(x^2+3x+2-5x-1-x-5)/((x+5)(x+1))`
`=(x^2-3x-4)/((x+5)(x+1))`
`=(x-4)/(x+5)`
c/ `P=A.B=-10/(x+5)` nguyên ``x \neq -5;x\neq-1;`x \neq -5;x\neq-1`
`<=>x+5 in Ư(10)={1;-1;5;-5;2;-2;10;-10}`
`<=> x in {-4;-6;0;-10;-3;-7;5;-15}` (TM)
