`(x+1)/x-3/(x-2)=(2-4x)/(x²-2x)(ĐKXĐ:``x`$\neq$ `0,x`$\neq$ `2)`
`⇔(x+1)/x-3/(x-2)=(2-4x)/[x(x-2)]`
`⇔[(x+1)(x-2)]/[x(x-2)]-(3x)/[x(x-2)]=(2-4x)/[x(x-2)]`
`⇒(x+1)(x-2)-3x=2-4x`
`⇔x²-2x+x-2-3x=2-4x`
`⇔x²-4x-2=2-4x`
`⇔x²-4x+4x=2+2`
`⇔x²=4`
`⇔`\(\left[ \begin{array}{l}x=2(Ko TM ĐKXĐ)\\x=-2(TM ĐKXĐ)\end{array} \right.\)
Vậy `S={-2}`
Câu 2:
`a)(x+4)/3-(x-3)/4≤(2x-1)/2+1`
`⇔[4(x+4)]/12-[3(x-3)]/12≤[6(2x-1)]/12+12/12`
`⇒4(x+4)-3(x-3)≤6(2x-1)+12`
`⇔4x+16-3x+9≤12x-6+12`
`⇔x+25≤12x+6`
`⇔x-12x≤6-25`
`⇔-11x≤-19`
`⇔x≥19/11`
Vậy `S={x|x≥19/11}`
Biểu diễn trên trục số:`(ảnh)`
`b)`
`|x-2|=4-3x`
`|x-2|=x-2` nếu `x≥2`
`2-x` nếu `x<2`
*Nếu `x≥2` ta có:
`x-2=4-3x`
`⇔x+3x=4+2`
`⇔4x=6`
`⇔x=6/4`
`⇔x=3/2(loại)`
*Nếu `x<2` ta có:
`2-x=4-3x`
`⇔-x+3x=4-2`
`⇔2x=2`
`⇔x=2:2`
`⇔x=1(TM)`
Vậy `S={1}`
