Đáp án + Giải thích các bước giải:
Nếu đề bài chứng minh thì :
`c)` `VT=1/(x-1)-(x^2-x)/(x^2+1)*(x/(x^2-2x+1)-1/(x^2-1))(xnepm1)`
`=(x^2+1)/[(x-1)(x^2+1)]-[(x^2-x)(x-1)]/[(x-1)(x^2+1)]*(x/[(x-1)^2] - 1/[(x-1)(x+1)])`
`=(x^2+1)/[(x-1)(x^2+1)]-(x^3-x^2-x^2+x)/[(x-1)(x^2+1)]*([x(x+1)]/[(x-1)^2(x+1)]-(x-1)/[(x-1)^2(x+1)])`
`=(x^2+1)/[(x-1)(x^2+1)]-(x^3-2x^2+x)/[(x-1)(x^2+1)]*(x^2+x-x+1)/[(x-1)^2(x+1)]`
`=(x^2+1)/[(x-1)(x^2+1)]-[x(x^2-2x+1)]/[(x-1)(x^2+1)]*(x^2+1)/[(x-1)^2(x+1)]`
`=(x^2+1)/[(x-1)(x^2+1)]-[x(x-1)^2]/[(x-1)(x^2+1)]*(x^2+1)/[(x-1)^2(x+1)]`
`=(x^2+1)/[(x-1)(x^2+1)]-x/[(x-1)(x+1)]`
`=[(x^2+1)(x+1)]/[(x-1)(x+1)(x^2+1)]-[x(x^2+1)]/[(x-1)(x+1)(x^2+1)]`
`=(x^3+x^2+x+1-x^3-x)/[(x-1)(x+1)(x^2+1)]=(x^2+1)/[(x^2-1)(x^2+1)]=1/(x^2-1)`.
`VP=1/(x^2-1)ne-1` `->` đề bài sai
+ Còn nếu tìm `x` thì `1/(x^2-1)=-1`
`<=>1/(x^2-1)=[-(x^2-1)]/(x^2-1)`
`=>1=1-x^2`
`<=>1-1+x^2=0`
`<=>x^2=0<=>x=0(tm)`
Vậy `S={0}`.
