`2x^2-5x+3=2x^2-2x-3x+3=2x(x-1)-3(x-1)=(x-1)(2x-3)`
`ĐKXĐ:x\ne1,x\ne3/2`
`a)B=({2x}/{2x^2-5x+3}-{5}/{2x-3}):(3+{2}/{1-x})`
`=[{2x}/{(x-1)(2x-3)}-{5}/{2x-3}]:[{3(1-x)}/{1-x}+{2}/{1-x}]`
`=[{2x}/{(x-1)(2x-3)}-{5(x-1)}/{(x-1)(2x-3)}]:{3(1-x)+2}/{1-x}`
`={2x-5(x-1)}/{(x-1)(2x-3)}:{3-3x+2}/{1-x}`
`={2x-5x+5}/{(x-1)(2x-3)}:{5-3x}/{1-x}`
`={5-3x}/{(x-1)(2x-3)}.{1-x}/{5-3x}`
`={1-x}/{(x-1)(2x-3)}`
`={-(x-1)}/{(x-1)(2x-3)}`
`={-1}/{2x-3}`
Vậy với `x\ne1,x\ne3/2` thì `B={-1}/{2x-3}`
`b,B=1/x^2`
`⇔{-1}/{2x-3}=1/x^2`
`⇔{-x^2}/{x^2(2x-3)}={2x-3}/{x^2(2x-3)}`
`⇒-x^2=2x-3`
`⇔-x^2-2x+3=0`
`⇔x^2+2x-3=0`
`⇔x^2+3x-x-3=0`
`⇔x(x+3)-(x+3)=0`
`⇔(x+3)(x-1)=0`
⇔\(\left[ \begin{array}{l}x+3=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3(TM)\\x=1(KTM)\end{array} \right.\)
Vậy với `x=-3` thì `B=1/x^2`
`c,B>0`
`⇔{-1}/{2x-3}>0`
Mà `-1<0`
`⇒B>0`
`⇔2x-3<0`
`⇔2x<3`
`⇔x<3/2`
Kết hợp ĐKXĐ:`x\ne1,x\ne3/2`
`⇒x<3/2,x\ne1`
Vậy với `x<3/2,x\ne1` thì `B>0`
