Đáp án:
`A_6=-2`
`A_7=x^2+2x+3`
Giải thích các bước giải:
`A_6=(x/(x^2-2x+1)-1/(x^2-1)).(2x-2x^3)/(x^2+1)+2/(x-1)(x`$\neq$ `±1)`
`=[x/(x-1)^2-1/[(x+1)(x-1)]].[-2x(x^2-1)]/(x^2+1)+2/(x-1)`
`=[[x(x+1)]/[(x+1)(x-1)^2]-(x-1)/[(x+1)(x-1)^2]].[-2x(x+1)(x-1)]/(x^2+1)+2/(x-1)`
`=[x(x+1)-(x-1)]/[(x+1)(x-1)^2].[-2x(x+1)(x-1)]/(x^2+1)+2/(x-1)`
`=(x^2+x-x+1)/[(x+1)(x-1)^2].[-2x(x+1)(x-1)]/(x^2+1)+2/(x-1)`
`=(x^2+1)/[(x+1)(x-1)^2].[-2x(x+1)(x-1)]/(x^2+1)+2/(x-1)`
`=(-2x)/(x-1)+2/(x-1)`
`=(-2x+2)/(x-1)`
`=[-2(x-1)]/(x-1)`
`=-2`
`A_7=(x+2)/(x-1).((x^3)/(x+1)+2)-(8x+7)/(x^2-1)(x`$\neq$ `±1)`
`=(x+2)/(x-1).[(x^3)/(x+1)+[2(x+1)]/(x+1)]-(8x+7)/[(x+1)(x-1)]`
`=(x+2)/(x-1).(x^3+2(x+1)]/(x+1)-(8x+7)/[(x+1)(x-1)]`
`=(x+2)/(x-1).(x^3+2x+2)/(x+1)-(8x+7)/[(x+1)(x-1)]`
`=[(x+2)(x^3+2x+2)]/[(x+1)(x-1)]-(8x+7)/[(x+1)(x-1)]`
`=[(x+2)(x^3+2x+2)-(8x+7)]/[(x+1)(x-1)]`
`=(x^4+2x^2+2x+2x^3+4x+4-8x-7)/[(x+1)(x-1)]`
`=[x^4+2x^3+2x^2+(2x+4x-8x)+(4-7)]/[(x+1)(x-1)]`
`=(x^4+2x^3+2x^2-2x-3)/[(x+1)(x-1)]`
`=(x^4+2x^3+3x^2-x^2-2x-3)/[(x+1)(x-1)]`
`=[(x^4+2x^3+3x^2)-(x^2+2x+3)]/[(x+1)(x-1)]`
`=[x^2(x^2+2x+3)-(x^2+2x+3)]/[(x+1)(x-1)]`
`=[(x^2-1)(x^2+2x+3)]/[(x+1)(x-1)]`
`=[(x+1)(x-1)(x^2+2x+3)]/[(x+1)(x-1)]`
`=x^2+2x+3`
