Đáp án:
`P=(x^3+17)/(x^2+x-2)-(2x-4)/(1-x)+3/(x+2)`
Điều kiện xác định:\(\begin{cases}x^2+x-2 \ne 0\\x+2 \ne 0\\x-1 \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}(x-1)(x+2) \ne 0\\x+2 \ne 0\\x-1 \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}x \ne 1\\x \ne -2\\\end{cases}\)
`P=(x^3+17)/((x+2)(x-1))+(2x-4)/(x-1)+3/(x+2)`
`P=(x^3+17)/((x+2)(x-1))+(2(x-2)(x+2))/(x-1)+(3(x-1))/((x-1)(x+2))`
`P=(x^3+17+2(x-2)(x+2)+3(x-1))/((x+2)(x-1))`
`P=(x^3+17+2x^2-8+3x-3)/((x+2)(x-1))`
`P=(x^3+2x^2+3x+6)/((x+2)(x-1))`
`P=(x^2(x+2)+3(x+2))/((x+2)(x-1))`
`P=((x+2)(x^2+3))/((x+2)(x-1))`
`P=(x^2+3)/(x-1).`
