Đáp án + Giải thích các bước giải:
`a//(3)/(2x-3)=(1)/(3-2x)+1` `(ĐK:x\ne(3)/(2))`
`⇔(3)/(2x-3)=-(1)/(2x-3)+(2x-3)/(2x-3)`
`⇒3=-1+2x-3`
`⇔-2x=-3-1-3`
`⇔-2x=-7`
`⇔x=(7)/(2)(TM)`
Vậy `S={(7)/(2)}`
`b//(2x)/(x^{2}-1)-(1)/(x+1)=2` `(ĐKXĐ:x\ne±1)`
`⇔(2x)/((x-1)(x+1))-(x-1)/((x+1)(x-1))=(2(x-1)(x+1))/((x-1)(x+1))`
`⇔2x-(x-1)=2(x-1)(x+1)`
`⇔2x-x+1=2x^{2}-2`
`⇔-2x^{2}+2x-x=-1-2`
`⇔-2x^{2}+x=-3`
`⇔-2x^{2}+x+3=0`
`⇔(-2x^{2}-2x)+(3x+3)=0`
`⇔-2x(x+1)+3(x+1)=0`
`⇔(x+1)(3-2x)=0`
`⇔` \(\left[ \begin{array}{l}x+1=0\\3-2x=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-1(KTM)\\x=\frac{3}{2}(TM)\end{array} \right.\)
Vậy `S={(3)/(2)}`