Đáp án + Giải thích các bước giải:
`a)` `(2x)/3+2<=(x-1)/2`
`<=>(2x)/3*6+2*6<=(x-1)/2*6`
`<=>4x+12<=3(x-1)`
`<=>4x+12<=3x-3`
`<=>4x+12-3x+3<=0`
`<=>x+15<=0`
`<=>x<=-15`
Vậy `S={x|x<=-15}`
`b)` `(2x-1)(x+3)=x^2-9`
`<=>(2x-1)(x+3)=(x-3)(x+3)`
`<=>(2x-1)(x+3)-(x-3)(x+3)=0`
`<=>(x+3)[(2x-1)-(x-3)]=0`
`<=>(x+3)(2x-1-x+3)=0`
`<=>(x+3)(x+2)=0`
`<=>`\(\left[ \begin{array}{l}x+3=0\\x+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-3\\x=-2\end{array} \right.\)
Vậy `S={-3;-2}`
`c)` `(x+2)/(x-1)-x/(x+1)=(2x-3)/(x^2-1)(xnepm1)`
`<=>[(x+2)(x+1)]/[(x-1)(x+1)]-[x(x-1)]/[(x-1)(x+1)]=(2x-3)/[(x-1)(x+1)]`
`=>(x+2)(x+1)-x(x-1)=2x-3`
`<=>x^2+x+2x+2-x^2+x-2x+3=0`
`<=>2x + 5 = 0`
`<=>2x=-5<=>x=-5/2(tm)`
Vậy `S={-5/2}`.
