`b)`
`A=(x+2)/(x+5)+(-5x-1)/(x^2+6x+5)-1/(x+1)`
`=(x+2)/(x+5)+(-5x-1)/((x+5)(x+1))-1/(x+1)`
`=((x+2)(x+1)+(-5x-1)-(x+5))/((x+5)(x+1))`
`=(x^2-3x-4)/((x+5)(x+1))`
`=((x+1)(x-4))/((x+1)(x+5))`
`=(x-4)/(x+5)`
``
`c)`
`P=A.B`
`\toP=(x-4)/(x+5) . (-10)/(x-4)`
`\toP=(-10)/(x+5)`
ĐKXĐ: `x\ne-5`
Để P nguyên `\toP\inZ\to(-10)/(x+5)\inZ`
`\to(-10)\vdots(x+5)`
`\to(x+5)\in Ư(-10)={+-1;+-2;+-5;+-10}`
Ta có bảng:
\begin{array}{|c|c|}\hline x+5&1&-1&2&-2&5&-5&10&-10\\\hline x&-4&-6&-3&-7&0&-10&5&-15\\\hline\end{array}
mà `x\ne-5\to x={--4;-6;-3;-7;-10;-15;0;5}`
