Đáp án + Giải thích các bước giải:
`a//(x-1)^{2}-1+x^{2}=(1-x)(x+3)`
`⇔(x-1)^{2}-(1-x^{2})-(1-x)(x+3)=0`
`⇔(x-1)^{2}-(1-x)(1+x)-(1-x)(x+3)=0`
`⇔(x-1)^{2}+(x-1)(x+1)+(x-1)(x+3)=0`
`⇔(x-1)(x-1+x+1+x+3)=0`
`⇔(x-1)(3x+3)=0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\3x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `S={±1}`
`b//x^{4}+x^{3}+x+1=0`
`⇔(x^{4}+x^{3})+(x+1)=0`
`⇔x^{3}(x+1)+(x+1)=0`
`⇔(x+1)(x^{3}+1)=0`
`⇔` \(\left[ \begin{array}{l}x+1=0\\x^{3}+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-1\\x^{3}=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-1\\x=-1\end{array} \right.\)
Vậy `S={-1}`
`c//(x^{2}-1)(x+2)(x-3)=(x-1)(x^{2}-4)(x+5)`
`⇔(x-1)(x+1)(x+2)(x-3)-(x-1)(x-2)(x+2)(x+5)=0`
`⇔(x-1)(x+2)[(x+1)(x-3)-(x-2)(x+5)]=0`
`⇔(x-1)(x+2)[x^{2}-2x-3-(x^{2}+3x-10)]=0`
`⇔(x-1)(x+2)(x^{2}-2x-3-x^{2}-3x+10)=0`
`⇔(x-1)(x+2)(-5x+7)=0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\x+2=0\\-5x+7=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=-2\\x=\frac{7}{5}\end{array} \right.\)
Vậy `S={1;-2;(7)/(5)}`
