C= 5xy – 3,5y² - 2 xy + 1,3 xy + 3x -2y

  =(5xy-2xy+1,3xy)-3,5y² +3x-2y

  =4,3xy-3,5y² +3x-2y

bậc của đa thức C là 2

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

\(P=\left(\dfrac{1}{3}x^2y-\dfrac{1}{3}x^2y\right)+\left(xy^2+\dfrac{1}{2}xy^2\right)-\left(xy+5xy\right)=\dfrac{3}{2}xy^2-6xy\)

\(A=5xy^2+xy-xy^2-\frac{1}{3}x^2y+2xy+x^2y+xy+6\)

\(A=\left(5xy^2-xy^2\right)+\left(xy+2xy+xy\right)+\left(-\frac{1}{3}x^2y+x^2y\right)+6\)

\(A=4xy^2+4xy+\frac{2}{3}x^2y+6\)

b) để A+B=0 => B là số đối của A 

\(\Rightarrow B=-4xy^2-4xy-\frac{2}{3}x^2y-6\)

c) Ta có \(A+C=-2xy+1\Leftrightarrow4xy^2+4xy+\frac{2}{3}x^2y+6+C=-2xy+1\)

\(\Leftrightarrow C=-2xy+1-4xy^2-4xy-\frac{2}{3}x^2y-6\)

\(\Leftrightarrow C=\left(-2xy-4xy\right)+\left(1-6\right)-4xy^2-\frac{2}{3}x^2y\)

\(\Leftrightarrow C=-6xy-5-4xy^2-\frac{2}{3}x^2y\)

\(a,-x^2y-2xy+2x^2y+5xy+2\\ =x^2y+3xy+2\\ b,-2xy+\dfrac{3}{2}xy^2+\dfrac{1}{2}xy^2+xy\\ =-xy+2xy^2\)

\(Q=5x^2y-3xy+\dfrac{1}{2}x^2y-xy+5xy-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}x-\dfrac{1}{4}\)

\(Q=\left(5-3+\dfrac{1}{2}+5-\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{1}{4}\right)+\left(x^2xx^2xxx\right)+\left(yyyyy\right)+\left(-x\right)\)

\(Q=\dfrac{97}{12}+x^{^{ }8}+y^5+\left(-x\right)\)

\(Q=\dfrac{97}{12}+x^7+y^5\)

a) 2x²yz + 4xy²z - 5x²yz + xy²z - xyz

= (2x²yz-5x²yz)+(4xy²z+xy²z)-xyz

= -3x²yz + 5xy²z - xyz

b) x³-5xy+3x³+xy-x²+\(\dfrac{1}{2}\)xy-x²

= (x³+3x³)+(xy-5xy+\(\dfrac{1}{2}\)xy)-(x²+x²)

= 4x³-\(\dfrac{7}{2}\)xy-2x²

x3 – 5xy + 3x3 + xy – x2 + 1/2.xy – x2

= (x3 + 3x3) + (xy + 1/2.xy – 5xy) – (x2 + x2)

= 4x3 - 7/2 xy – 2x2

A=(\(\dfrac{4}{9}x^4y^2\)).\(\dfrac{2}{5}xy^3\)

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

=\(\dfrac{8}{45}x^5y^5\)

\(P=\dfrac{1}{3}x^2y+xy^2-xy+\dfrac{1}{2}xy^2-5xy-\dfrac{1}{3}x^2y=\dfrac{3}{2}xy^2-6xy\)

Thay x = 2 ; y = 1 ta được 

\(\dfrac{3}{2}.2.1-6.2.1=3-12=-9\)