`1)`
`a)(x-3)(x+4)-2(3x-2)=(x-4)^2`
`⇔x^2+4x-3x-12-6x+4=x^2-8x+16`
`⇔x^2-5x-8=x^2-8x+16`
`⇔x^2-x^2-5x+8x=16+8`
`⇔3x=24`
`⇔x=8`
Vậy `S={8}`
`b)(2x-1)(3x+2)-4x^2+1=0`
`⇔(2x-1)(3x+2)-(4x^2-1)=0`
`⇔(2x-1)(3x+2)-(2x-1)(2x+1)=0`
`⇔(2x-1)(3x+2-2x-1)=0`
`⇔(2x-1)(x+1)=0`
`1)2x-1=0⇔x=1/2`
`2)x+1=0⇔x=-1`
Vậy `S={1/2;-1}`
`c)ĐKXĐ:x`$\neq$`0;x`$\neq$`2`
Ta có: `(x+2)/(x-2)-2/(x^2-2x)=1/x`
`⇔(x^2+2x)/(x^2-2x)-2/(x^2-2x)=(x-2)/(x^2-2x)`
`⇒x^2+2x-2=x-2`
`⇔x^2+2x-x=-2+2`
`⇔x^2+x=0`
`⇔x(x+1)=0`
`1)x=0(loại)`
`2)x+1=0⇔x=-1(TM)`
Vậy `S={-1}`
`d)|2-x|=3x+5`
`|2-x|=`$\left\{\begin{matrix} 2-x nếu x\leq2 \\\\x-2 nếu x>2 \end{matrix}\right.$
Xét TH:`x`$\leq$ `2`
`2-x=3x+5`
`⇔-4x=3`
`⇔x=-3/4(TM)`
Xét TH:`x>2`
`x-2=3x+5`
`⇔-2x=7`
`⇔x=-7/2(loại)`
Vậy pt đã cho có nghiệm:`x=-3/4`
`2.`
`a)(9x-2)/3>(5x+9)/2`
`⇔(18x-4)/6>(15x+27)/6`
`⇔18x-4>15x+27`
`⇔3x>31`
`⇔x>(31)/3`
Vậy BPT có nghiệm `x>(31)/3`
Biểu diễn:ảnh
`b)(3x+1)/4-(5x+1)/6<(x+4)/3`
`⇔(9x+3)/(12)-(10x+2)/(12)<(4x+16)/(12)`
`⇔9x+3-10x-2<4x+16`
`⇔-x+1<4x+16`
`⇔-5x<15`
`⇔x> -3`
BPT có nghiệm `x> -3`
Biểu diễn:ảnh
