Question

(\frac{-4x^{2}(3x - 3y + 2)}{2x}-\frac{6x^{2}(3 - 2x + y)}{-x})

Explanation:

Step1: Simplify first fraction

Divide each term by $2x$:
$\frac{-4x^2(3x-3y+2)}{2x} = -2x(3x-3y+2) = -6x^2 + 6xy - 4x$

Step2: Simplify second fraction

Divide each term by $-x$:
$\frac{6x^2(3-2x+y)}{-x} = -6x(3-2x+y) = -18x + 12x^2 - 6xy$

Step3: Subtract the two results

Subtract the simplified second fraction from the first:
$(-6x^2 + 6xy - 4x) - (-18x + 12x^2 - 6xy)$

Step4: Distribute the negative sign

$ -6x^2 + 6xy - 4x + 18x - 12x^2 + 6xy$

Step5: Combine like terms

Combine $x^2$ terms: $-6x^2 -12x^2 = -18x^2$
Combine $xy$ terms: $6xy + 6xy = 12xy$
Combine $x$ terms: $-4x + 18x = 14x$

Answer:

$-18x^2 + 12xy + 14x$