Đáp án:
Giải thích các bước giải:
`a)(x+2)-x(x+2)=0`
`<=>(1-x)(x+2)=0`
`<=>` \(\left[ \begin{array}{l}1-x=0\\x+2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
Vậy `S={-2;1}`
`b)` `(2x+3)/4-x=(1-9x)/12`
`<=>(3(2x+3)-12x)/12=(1-9x)/12`
`=>6x+9-12x=1-9x`
`<=>-6x+9=1-9x`
`<=>-6x+9x=1-9`
`<=>3x=-8`
`<=>x=-8/3`
Vậy `S={-8/3}`
`c)` `(x+3)/(x-3)-3/(x(x-3))=1/x` (đk: `x\ne0;x\ne3`)
`<=>(x(x+3)-3)/(x(x-3))=(x-3)/(x(x-3))`
`=>x^2+3x-3=x-3`
`<=>x^2+3x-x-3+3=0`
`<=>x^2+2x=0`
`<=>x(x+2)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x+2=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0(ktmđk)\\x=-2(tmđk)\end{array} \right.\)
Vậy `S={-2}`
