`B=[x^2+7|-x|+12]/[(2x-4)^2-(1-x)^2]` `(x < 0)`
`B=[x^2-7x+12]/[(2x-4-1+x)(2x-4+1-x)]`
`B=[x^2-3x-4x+12]/[(3x-5)(x-3)]`
`B=[(x-3)(x-4)]/[(3x-5)(x-3)]`
`B=[x-4]/[3x-5]`
`->bb B`
\(B=\dfrac{x^2-7x+12}{\left(2x-4-1+x\right)\left(2x-4+1-x\right)}=\dfrac{\left(x-3\right)\left(x-4\right)}{\left(3x-5\right)\left(x-3\right)}=\dfrac{x-4}{3x-5}\)
\(B=\dfrac{x^2-7x+12}{\left(2x-4\right)^2-\left(1-x\right)^2}=\dfrac{x^2-3x-4x+12}{4x^2-16x+16-1+2x-x^2}=\dfrac{x\left(x-3\right)-4\left(x-3\right)}{3x^2-14x+15}=\dfrac{\left(x-4\right)\left(x-3\right)}{3x^2-9x-5x+15}=\dfrac{\left(x-4\right)\left(x-3\right)}{3x\left(x-3\right)-5\left(x-3\right)}=\dfrac{\left(x-4\right)\left(x-3\right)}{\left(3x-5\right)\left(x-3\right)}=\dfrac{x-4}{3x-5}\)
