a,$\frac{5}{2x^2y}+\frac{3}{5xy^2}+\frac{x}{y^3}$52x2y +35xy2 +xy3 

b,\(\frac{x+1}{2x+6}+\frac{2x+3}{x\left(x+3\right)}\)

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\(=\frac{16+x}{x^2-2x}-\frac{18}{x^2-2x}\)

\(=\frac{16+x-18}{x\left(x-2\right)}\)

\(=\frac{-2+x}{x\left(x-2\right)}\)

a) \(\frac{16+x}{x^2-2x}+\frac{18}{2x-x^2}=\frac{16+x-18}{x^2-2x}=\frac{x-2}{x\left(x-2\right)}=\frac{1}{x}\)

b) \(\frac{2y}{2x^2-xy}+\frac{4x}{xy-2x^2}=\frac{2y-4x}{2x^2-xy}=\frac{-2\left(2x-y\right)}{x\left(2x-y\right)}=\frac{-2}{x}\)

c) \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}=\frac{4-x^2+2x^2-2x+5-4x}{x-3}=\frac{x^2-6x+9}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)

a)

\(\frac{16+x}{x^2-2x}+\frac{18}{2x-x^2}=\frac{16+x}{x^2-2x}-\frac{18}{x^2-2x}\)

\(=\frac{16+x-18}{x\left(x-2\right)}=\frac{x-2}{x\left(x-2\right)}=\frac{1}{x}\)

b)   \(\frac{2y}{2x^2-xy}+\frac{4x}{xy-2x^2}=\frac{2y}{2x^2-xy}-\frac{4x}{2x^2-xy}\)

\(=\frac{2y-4x}{x\left(2x-y\right)}=\frac{2\left(y-2x\right)}{x\left(2x-y\right)}=-\frac{2}{x}\)

c)  \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}\)\(=\frac{4-x^2}{x-3}-\frac{2x-2x^2}{x-3}+\frac{5-4x}{x-3}\)

\(=\frac{4-x^2-2x+2x^2+5-4x}{x-3}\)\(=\frac{x^2-6x+9}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)

a) \(\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}\)

\(=\frac{2x-7}{10x-4}-\frac{-\left(3x+5\right)}{-\left(4-10x\right)}\)

\(=\frac{2x-7}{10x-4}-\frac{5-3x}{10x-4}\)

\(=\frac{2x-7-\left(5-3x\right)}{10x-4}\)

\(=\frac{2x-7-5+3x}{10x-4}\)

\(=\frac{5x-12}{10x-4}\)

`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

`b, x(x-2y) - y(y^2-2x)`

`= x^2 - 2xy - y^3 + 2xy`

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)

\(=x^2-2xy-y^3+2xy\)

\(=x^2-y^3\)

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)

Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)

\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)

\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)

Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)

\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)

\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)

\(P=\frac{1}{3}x^2y+xy^2-xy+\frac{1}{2}xy^2-5xy-\frac{1}{3}x^2y\)

\(=\left(\frac{1}{3}x^2y-\frac{1}{3}x^2y\right)+\left(xy^2+\frac{1}{2}xy^2\right)-\left(xy+5xy\right)\)

\(=0+xy^2\left(1+\frac{1}{2}\right)-xy\left(1+5\right)\)

\(=\frac{3}{2}xy^2-6xy\)

Thay x = 0,5 và y = 1 vào đa thức trên ta có:

\(\frac{3}{2}.0,5.1^2-6.0,5.1\)

\(=\frac{3}{2}.\frac{1}{2}.1-6.\frac{1}{2}.1\)

\(=\frac{3}{4}-\frac{6}{2}=\frac{3}{4}-\frac{12}{4}=-\frac{9}{4}\)

P/s: Ko chắc!

P=\(\frac{1}{3}\)\(^{x^2}\)y+\(^{xy^2}\)−xy+\(\frac{1}{2}\)\(^{xy^2}\)−5xy−\(\frac{1}{2}\)\(^{xy^2}\)

P=(\(\frac{1}{3}\)−\(\frac{1}{3}\))+(\(^{xy^2}\)+\(\frac{1}{2}\)\(^{xy^2}\))+(−xy−5xy)

P=0+\(\frac{3}{2}\)\(^{xy^2}\)+(−6xy)=\(\frac{3}{2}\)\(^{xy^2}\)−6xy

Thay x=0,5;y=1 vào P ta có:

\(\frac{3}{2}\).0,5.\(^{1^2}\)−6.0,5.1=\(\frac{3}{4}\)−3=\(\frac{3}{4}\)−\(\frac{12}{4}\)=\(\frac{-9}{4}\)=−2,25

Vậy P=\(\frac{-9}{4}\)=-2,25