Đáp án:
`S={0}`
Giải thích các bước giải:
`b)(-7x²+4)/(x³+1)=5/(x²-x+1)-1/(x+1)(ĐKXĐ:x`$\neq$ `-1)`
`⇔(-7x²+4)/[(x+1)(x²-x+1)]=5/(x²-x+1)-1/(x+1)`
`⇔(-7x²+4)/[(x+1)(x²-x+1)]=[5(x+1)]/[(x+1)(x²-x+1)]-(x²-x+1)/[(x+1)(x²-x+1)]`
`⇒-7x²+4=5(x+1)-(x²-x+1)`
`⇔-7x²+4=5x+5-x²+x-1`
`⇔-7x²-5x+x²-x=5-1-4`
`⇔-6x²-6x=0`
`⇔-6x(x+1)=0`
`⇔`\(\left[ \begin{array}{l}-6x=0\\x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0(TMĐKXĐ)\\x=-1(Ko TMĐKXĐ)\end{array} \right.\)
Vậy `S={0}`
