Đáp án + Giải thích các bước giải:
`a)` `1/(x-1)-(3x^2)/(x^3-1)=(2x)/(x^2+x+1) ( x ne 1)`
`<=>1/(x-1)-(3x^2)/[(x-1)(x^2+x+1)]=(2x)/(x^2+x+1)`
`<=>(x^2+x+1)/[(x-1)(x^2+x+1)]-(3x^2)/[(x-1)(x^2+x+1)]=[2x(x-1)]/[(x-1)(x^2+x+1)]`
`=>x^2+x+1-3x^2=2x^2-2x`
`<=>x^2+x+1-3x^2-2x^2+2x=0`
`<=>-4x^2+3x+1=0`
`<=>-4x^2+4x-x+1=0`
`<=>-4x(x-1)-(x-1)=0`
`<=>(x-1)(-4x-1)=0`
`<=>`\(\left[ \begin{array}{l}x=1(ktm)\\x=-\dfrac{1}{4}(tm)\end{array} \right.\)
Vậy `S={-1/4}`
`b)` `1/(x^2-2x+4)+1/(x+2)=12/(x^3+8)(xne-2)`
`<=>(x+2)/[(x+2)(x^2-2x+4)]+(x^2-2x+4)/[(x+2)(x^2-2x+4)]=12/[(x+2)(x^2-2x+4)]`
`=>x+2+x^2-2x+4=12`
`<=>x^2-x+6=12`
`<=>x^2-x-6=0`
`<=>x^2+2x-3x-6=0`
`<=>x(x+2)-3(x+2)=0`
`<=>(x+2)(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=-2(ktm)\\x=3(tm)\end{array} \right.\)
Vậy `S={3}`.
