Đáp án:
`c)`
ĐKXĐ : `x ne +-2`
`(x-2)/(x+2)+3/(x-2)=(x^2-11)/(x^2-4)`
`<=> ((x-2)^2+3(x+2))/(x^2-4)=(x^2-11)/(x^2-4)`
`=> (x-2)^2+3(x+2)=x^2-11`
`<=> x^2-4x+4+3x+6-x^2+11=0`
`<=> -x=-21`
`<=> x=21 \ \ (tm)`
Vậy `S={21}`
`d)`
ĐKXĐ : `x ne -1;2`
`2/(x+1)-1/(x-2)=(3x-11)/((x+1)(x-2))`
`<=> (2(x-2)-(x+1))/((x+1)(x-2))=(3x-11)/((x+1)(x-2))`
`=> 2(x-2)-(x+1)=3x-11`
`<=> 2x-4-x-1-3x+11=0`
`<=> -2x=-6`
`<=> x=3 \ \ (tm)`
Vậy `S={3}`
`e)`
ĐKXĐ : `x ne 0;2`
`(x+2)/(x-2)-1/x=2/(x^2-2x)`
`<=> ((x+2)x-(x-2))/(x(x-2))=2/(x(x-2))`
`=> (x+2)x-(x-2)=2`
`<=> x^2+2x-x+2=2`
`<=> x^2+x=0`
`<=> x.(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0 \ \ \rm (ktm)\\x=-1 \ \ \rm (tm)\end{array} \right.\)
Vậy `S={-1}`
