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

Bậc:4

\(\left(x^2-2xy+y^2\right)+81\)

em.........

bó tay 

Ta có :

 \(A\left(x\right)-B\left(x\right)=3x^3y^4-2xy^2-5xy-1-3x^3y^4+2xy^2+xy+4\)

\(=-4xy+3\)bậc 2 

\(A\left(x\right)+B\left(x\right)=3x^3y^4-2xy^2-5xy-1+3x^3y^4-2xy^2-xy-4\)

\(=6x^3y^4-4xy^2-6xy-5\)bậc 7 

\(a,=6y\left(2x^2-3xy-5y^2\right)\\ =6y\left(2x^2+2xy-5xy-5y^2\right)\\ =6y\left(x+y\right)\left(2x-5y\right)\\ b,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ c,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)\\ =\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ d,=\left(a^2+3b\right)^2-1=\left(a^2+3b+1\right)\left(a^2+3b-1\right)\\ e,=\left(2x-5\right)\left(2x+5\right)-\left(2x+7\right)\left(2x-5\right)\\ =\left(2x-5\right)\left(2x+5-2x-7\right)\\ =-2\left(2x-5\right)\\ f,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ g,=x^3-x-6x-6\\ =x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x-6\right)\\ =\left(x+1\right)\left(x^2-3x+2x-6\right)\\ =\left(x+1\right)\left(x-3\right)\left(x+2\right)\\ l,=x^4+4x^2+4-4x^2\\ =\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\\ h,=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)

=(x-y)²-9²

=(x-y-9)(x-y+9)

b) \(5x^3+10x^2y+5xy^2=2\left(x^3+2x^2y+xy^2\right)\)

\(=2\left(x^3+x^2y+x^2y+xy^2\right)=2\left[x^2\left(x+y\right)+xy\left(x+y\right)\right]\)

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

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

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

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

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

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

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

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

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

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²