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

`a, x/(x+3) + (2-x)/(x+3) = (x+2-x)/(x+3) = 2/(x+3)`

`b, (x^2y)/(x-y) - (xy^2)/(x-y) = (x^2y-xy^2)/(x-y) = (xy(x-y))/(x-y)= xy`

`c, (2x)/(2x-y) - (y)/(2x-y)`

`= (2x-y)/(2x-y) = 1`

a: \(=\dfrac{3b+4a}{6ab}\)

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

c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)

d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)

e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)

\(a,=\dfrac{4y.5x^3}{3x^2.2y^3}=\dfrac{20x^3y}{6x^2y^3}=\dfrac{10x}{3y^2}\\ b,=\dfrac{\left(x-1\right)^2.x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2.x.\left(x+1\right)}{\left(x-1\right)^2.\left(x+1\right)}=x\)

\(c,=\dfrac{x\left(2+x\right).3\left(x^3+1\right)}{\left(x^2-x+1\right).3.\left(x+2\right)}=\dfrac{3x.\left(x+2\right).\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right).3\left(x+2\right)}=x\left(x+1\right)\)

`a, a/(a-3) - 3/(a+3) = (a(a+3) - 3(a-3))/(a^2-9)`

`= (a^2+9)/(a^2-9)`

`b, 1/(2x) + 2/x^2 = x/(2x^2) + 4/(2x^2) = (x+4)/(2x^2)`

`c, 4/(x^2-1) - 2/(x^2+x) = (4x)/(x(x-1)(x+1)) - (2(x-1))/(x(x+1)(x-1))`

`= (2x+2)/(x(x-1)(x+1)`

`= 2/(x(x-1))`

\(a,=\dfrac{5x}{4y^3}\times\left(\dfrac{-20y}{x^4}\right)=\dfrac{-100xy}{4x^4y^3}=\dfrac{-25}{x^3y^2}\\ b,=\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+4\right)}\times\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)

\(c,=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\times\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x+3\right)^2.\left(x^2+2x+4\right)}\)

a) \(\dfrac{5x}{4y^3}:\left(-\dfrac{x^4}{20y}\right)=\dfrac{5x}{4y^3}\cdot\left(-\dfrac{20y}{x^4}\right)=\dfrac{5\cdot-5}{y^2\cdot x^3}=\dfrac{-25}{x^3y^2}\)

b) \(\dfrac{x^2-16}{x+4}:\dfrac{2x-8}{x}=\left(x-4\right)\cdot\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)

c) \(\dfrac{2x+6}{x^3-8}:\dfrac{\left(x+3\right)^3}{2x-4}=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x^2+2x+4\right)\left(x+3\right)^2}\)

a) \(\dfrac{a-1}{a+1}+\dfrac{3-a}{a+1}\)

\(=\dfrac{a-1+3-a}{a+1}\)

\(=\dfrac{2}{a+1}\)

b) \(\dfrac{b}{a-b}+\dfrac{a}{b-a}\)

\(=\dfrac{b}{a-b}+\dfrac{-a}{a-b}\)

\(=\dfrac{b-a}{a-b}\)

\(=-1\)

c) \(\dfrac{\left(a+b\right)^2}{ab}-\dfrac{\left(a-b\right)^2}{ab}\)

\(=\dfrac{\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)+\left(a-b\right)\right]}{ab}\)

\(=\dfrac{4ab}{ab}\)

\(=4\)

`a, (a-1)/(a+1) + (3-a)/(a+1)`

`= (a-1+3-a)/(a+1)`

`=2/(a+1)`

`b, b/(a-b) + a/(b-a)`

`=  b/(a-b) - a/(a-b)`

`= (b-a)/(a-b)`

`c, (a+b)^2/(ab) -(a-b)^2/(ab)`

`=(a^2+2ab+b^2-a^2+2ab-b^2)/(ab)`

`= (4ab)/(ab)`

`a)`

`4x^3 * (-6x^3y)`

`= 4*(-6) * (x^3*x^3) * y`

`= -24x^6y`

`b)`

`(-2y)*(-5xy^2)`

`= (-2)*(-5)*x*(y*y^2)`

`= 10xy^3`

`c)`

`(-2a)^3 * (2ab)^2`

`= (-8a^3) * (4a^2b^2)`

`= (-8*4)*(a^3*a^2)*b^2`

`= -32a^5b^2`

a) \(4x^3\cdot\left(-6x^3y\right)\)

\(=\left(4\cdot-6\right)\cdot\left(x^3\cdot x^3\right)\cdot y\)

\(=-24x^6y\)

b) \(\left(-2y\right)\cdot\left(-5xy^2\right)\)

\(=\left(-2\cdot-5\right)\cdot\left(y\cdot y^2\right)\cdot x\)

\(=10xy^3\)

c) \(\left(-2a\right)^3\cdot\left(2ab\right)^2\)

\(=-8a^3\cdot4a^2b^2\)

\(=\left(-8\cdot4\right)\cdot\left(a^3\cdot a^2\right)\cdot b^2\)

\(=-32a^5b^2\)

a) \({x^2} + \dfrac{1}{4}{x^2} - 5{x^2} = (1 + \dfrac{1}{4} - 5){x^2} =  - \dfrac{{15}}{4}{x^2}\);

b) \({y^4} + 6{y^4} - \dfrac{2}{5}{y^4} = (1 + 6 - \dfrac{2}{5}){y^4} = \dfrac{{33}}{5}{y^4}\).