Đáp án + Giải thích các bước giải:
`a//(1-x)/(x+1)+3=(2x+3)/(x+1)` `(ĐKXĐ:x\ne-1)`
`⇔(1-x)/(x+1)+(3(x+1))/(x+1)=(2x+3)/(x+1)`
`⇔1-x+3(x+1)=2x+3`
`⇔1-x+3x+3=2x+3`
`⇔-x+3x-2x=-1-3+3`
`⇔0x=-1` ( Vô nghiệm )
Vậy phương trình vô nghiệm
`b//(x-3)/(x-2)+(x-2)/(x-4)=-1` `(ĐKXĐ:x\ne{2;4})`
`⇔((x-3)(x-4))/((x-2)(x-4))+((x-2)^{2})/((x-4)(x-2))=-((x-4)(x-2))/((x-4)(x-2))`
`⇔(x-3)(x-4)+(x-2)^{2}=-(x-4)(x-2)`
`⇔x^{2}-7x+12+x^{2}-4x+4=-(x^{2}-6x+8)`
`⇔2x^{2}-11x+16=-x^{2}+6x-8`
`⇔2x^{2}+x^{2}-11x-6x+16+8=0`
`⇔3x^{2}-17x+24=0`
`⇔(3x^{2}-9x)-(8x-24)=0`
`⇔3x(x-3)-8(x-3)=0`
`⇔(x-3)(3x-8)=0`
`⇔` \(\left[ \begin{array}{l}x-3=0\\3x-8=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3(TM)\\x=\frac{8}{3}(TM)\end{array} \right.\)
Vậy `S={3;(8)/(3)}`
