Đáp án:
`1.`
`a) \ \ 5x+1=4x-2`
`<=> 5x-4x=-2-1`
`<=> x=-3`
Vậy `S={-3}`
`b) \ \ (x-1)/x+(1-2x)/(x(x+1))=1/(x+1)`
ĐKXĐ : `x ne 0;-1`
`<=> ((x-1)(x+1)+(1-2x))/(x(x+1))=x/(x(x+1))`
`=> (x-1)(x+1)+(1-2x)=x`
`<=> x^2-1+1-2x-x=0`
`<=> x^2-3x=0`
`<=> x.(x-3)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0 \ \rm (ktm)\\x=3 \ \rm (tm)\end{array} \right.\)
Vậy `S={3}`
`2.`
`a) \ \ a>b`
`<=> 2a>2b`
`<=> 2a-3>2b-3`
`b) \ \ (2-x)/3 < (3-2x)/5`
`<=> 5(2-x)<3(3-2x)`
`<=> 10-5x<9-6x`
`<=> -5x+6x<9-10`
`<=> x<-1`
Vậy `S={x|x<-1}`
