Đáp án:
`a, S={-3;1/2}`
`b, S=∅`
Giải thích các bước giải:
`a, (x+3)(2x-1)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\2x-1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-3\\x=1/2\end{array} \right.\)
Vậy `S={-3;1/2}`
`b, \frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x²-1}`
(ĐKXĐ: `x≠±1`)
`=> \frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{(x-1)(x+1)}`
`<=> \frac{(x+1)²-(x-1)²}{(x-1)(x+1)}=\frac{4}{(x-1)(x+1)}`
`<=> (x+1-(x-1))(x+1+x-1)=4`
`<=>( x+1-x+1).2x=4`
`<=> 2.2x=4`
`<=>4x=4`
`<=> x=1` (Không TMĐK)
Vậy `S=∅`
