Đáp án+Giải thích các bước giải:
`a,1/x+2/(x-2)=0 ĐKXĐ:x \ne 0;x\ne 2`
`⇔(x-2+2x)/(x(x-2))=0`
`⇔(3x-2)/(x(x-2))=0`
`⇒3x-2=0`
`⇔x=2/3∈ĐKXĐ`
Vậy `S={2/3}`
`b,(x-1)/(x+2)-x/(x-2)=(5x-2)/(4-x^2)` ĐKXĐ `: x\ne ±2`
`⇔(x-1)/(x+2)-x/(x-2)-(5x-2)/(4-x^2)=0`
`⇔(x-1)/(x+2)-x/(x-2)+(5x-2)/((x-2)(x+2))=0`
`⇔((x-1)(x-2)-x(x+2)+5x-2)/((x-2)(x+2))=0`
`⇒x^2-2x-x+2-x^2-2x+5x-2=0`
`⇔0x=0`
⇒phương trình vô số nghiệm
Vậy `S ∈ R`
`c,1+1/(2+x)=12/(x^3+8) ĐKXĐ: x\ne -2`
`⇔1+1/(x+2)-12/((x+2)(x^2-2x+4))=0`
`⇔(x^3+8+x^2-2x+4-12)/((x+2)(x^2-2x+4))=0`
`⇒x^3+x^2-2x=0`
`⇔x(x^2+x-2)=0`
`⇔x(x-1)(x+2)=0`
⇔\(\left[ \begin{array}{l}x=0∈ĐKXĐ\\x=1∈ĐKXĐ\\x=-2∉ĐKXĐ\end{array} \right.\)
Vậy `S={0;1}`
