Đáp án:
`c)(x+1)/(x^2+x+1)`
`d)x/(x+1)`
Giải thích các bước giải:
`c)(3x²+5x+1)/(x³-1)-(1-x)/(x²+x+1)-3/(x-1)(x\ne1)`
`=(3x²+5x+1)/[(x-1)(x²+x+1)]+(x-1)/(x²+x+1)-3/(x-1)`
`=(3x²+5x+1)/[(x-1)(x²+x+1)]+[(x-1)²]/[(x-1)(x²+x+1)]-[3(x²+x+1)]/[(x-1)(x²+x+1)]`
`=[(3x²+5x+1)+(x-1)²-3(x²+x+1)]/[(x-1)(x²+x+1)]`
`=(3x^2+5x+1+x^2-2x+1-3x^2-3x-3)/[(x-1)(x^2+x+1)]`
`=[(3x^2+x^2-3x^2)+(5x-2x-3x)+(1+1-3)]/[(x-1)(x^2+x+1)]`
`=(x^2-1)/[(x-1)(x^2+x+1)]`
`=[(x+1)(x-1)]/[(x-1)(x^2+x+1)]`
`=(x+1)/(x^2+x+1)`
`d)1/(x^2-x+1)+1-(x^2+2)/(x^3+1)(x\ne-1)`
`=1/(x^2-x+1)+1-(x^2+2)/[(x+1)(x^2-x+1)]`
`=(x+1)/[(x+1)(x^2-x+1)]+[(x+1)(x^2-x+1)]/[(x+1)(x^2-x+1)]-(x^2+2)/[(x+1)(x^2-x+1)]`
`=[(x+1)+(x+1)(x^2-x+1)-(x^2+2)]/[(x+1)(x^2-x+1)]`
`=(x+1+x^3+1-x^2-2)/[(x+1)(x^2-x+1)]`
`=(x^3-x^2+x)/[(x+1)(x^2-x+1)]`
`=[x(x^2-x+1)]/[(x+1)(x^2-x+1)]`
`=x/(x+1)`
