\(a,=\dfrac{x+3+4x-3}{xy}=\dfrac{5x}{xy}=\dfrac{5}{y}\\ b,=\dfrac{x-2+4x-3}{x-1}=\dfrac{5\left(x-1\right)}{x-1}=5\\ c,=\dfrac{x^2+4-6x+5}{x-3}=\dfrac{\left(x-3\right)^2}{x-3}=x-3\\ d,=\dfrac{4-x^2+2x^2-2x+5-4x}{x-3}=\dfrac{\left(x-3\right)^2}{x-3}=x-3\)

a) \(\dfrac{5xy^2-x^2y+4xy^2+5x^2y}{3xy}=\dfrac{9xy^2+4x^2y}{3xy}=\dfrac{xy\left(9y+4x\right)}{3xy}=\dfrac{9y+4x}{3}\)

b) \(\dfrac{x+3+x-1+x+4}{x+2}=\dfrac{3x+6}{x+2}=\dfrac{3\left(x+2\right)}{x+2}=3\)

\(\dfrac{5xy^2-x^2y}{3xy}+\dfrac{4xy^2+5x^2y}{3xy}=\dfrac{5xy^2-x^2y+4xy^2+5x^2y}{3xy}=\dfrac{9xy^2+4x^2y}{3xy}=\dfrac{xy\left(9y+4x\right)}{3xy}=\dfrac{9y+4x}{3}\)

\(\dfrac{x+3}{x+2}+\dfrac{x-1}{x+2}+\dfrac{x+4}{x+2}=\dfrac{x+3+x-1+x+4}{x+2}=\dfrac{3x+6}{x+2}=\dfrac{3\left(x+2\right)}{x+2}=3\)

b: \(=\dfrac{x+3+x-1+x+4}{x+2}=\dfrac{3x+6}{x+2}=3\)

\(a,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{3x^2}\\ b,=\dfrac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\\ d,=\dfrac{1}{\left(x+3\right)^2}-\dfrac{1}{\left(x-3\right)^2}+\dfrac{x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\dfrac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)

a)   (x+2)² + (x + 1)(x - 1) - x(2x + 5)

    =   x² + 4x +4 + x² - 1 - 2x² - 5x

    = 3 - x

b)    (35x⁶y⁷ - 25x³y⁵): 5x³y⁵ - \(\dfrac{1}{3}x^3y^2\)

    = 7x³y² - 5 - \(\dfrac{1}{3}x^3y^2\)

     = \(\dfrac{20}{3}\)x³y² -5

c)   \(\dfrac{x-3}{x+3}+\dfrac{3}{3-x}+\dfrac{6x}{x^2-9}\)

= \(\dfrac{x-3}{x+3}-\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}\)

= \(\dfrac{\left(x-3\right)^2-3\left(x+3\right)+6x}{x^2-9}\)

= \(\dfrac{x^2-6x+9-3x-9+6x}{x^2-9}\)

= \(\dfrac{x^2-3x}{x^2-9}\)

= \(\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{x}{x+3}\)

a: =(x-y)(x+y)-2(x-y)

=(x-y)(x+y-2)

\(=x^2+4x+4+x^2-1-2x^2-5x=-x+3\)

g: \(=-2x^3+3x-4\)

f. \(4x^3-4x^2+4x-\left(4x-1\right)\left(x^2-x\right)\)
g\(-2x^3+3x-4\)

\(=4x^3-4x^2+4x-\left(4x^3-4x^2-x^2+x\right)\\ =4x^3-4x^2+4x-4x^3+5x^2-x=x^2+3x\)