Đáp án+Giải thích các bước giải:
9.
`(x-1)/(x+1) - (x^2 + x-2)/(x+1) = (x+1)/(x-1) -x-2` `(ĐKXĐ: x ne +-1)`
`<=> (x-1)/(x+1) - (x^2 + x-2)/(x+1) - (x+1)/(x-1) + x =-2`
`<=> ((x-1)^2 - (x-1) (x^2 + x -2) - (x-1)^2 + x(x+1)(x-1))/((x+1)(x-1)) =-2`
`<=> ((x-1)^2 - (x^3 + x^2 - 2x - x^2 - x +2) - (x-1)^2 + x(x^2-1))/((x+1)(x-1)) =-2`
`<=> ((x-1)^2 -(x^3 - 3x +2) + (x+1)^2 + x^3 -x)/((x+1)(x-1)) =-2`
`<=> (-2*2x - x^3 + 3x - 2 + x^3 -x)/((x+1)(x-1)) =-2`
`<=> (-2x-2)/((x+1)(x-1)) =-2`
`<=> (-2(x+1))/(x-1) =-2`
`<=>2= (x-1)*2`
`<=> 1= x-1`
`<=> -x=-2`
`<=> x=2`
Vậy `S ={2}`
10.
`x^4 + 3x^3 + 4x^2 + 3x +1 =0`
`<=> x^4 + x^3 + 2x^3 + 2x^2 + 2x^2 + 2x + x +1 =0`
`<=> x^3(x+1) + 2x^2 (x+1) + 2x(x+1) + (x+1) =0`
`<=> (x+1)(x^3 + 2x^2 + 2x +1)=0`
`<=> (x+1)[(x+1)(x^2 -x+1)+2x(x+1)]=0`
`<=> (x+1)(x+1)(x^2 +x +1) =0`
`<=>` \(\left[ \begin{array}{l}(x+1)^2 =0\\\\x^2 +x +1 =0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x-1\\\\x \notin \mathbb{R}\end{array} \right.\)
Vậy `S = {-1}`
