a) \(=\dfrac{5}{2}xy+\dfrac{3}{2}x^2\)

b) \(=-3y-2x^2\)

c) \(=x-\dfrac{3}{5}x^2z\)

d) \(=-\dfrac{5}{2}x^2z+4x^3\)

a)M=3x²y-2xy²+2x²y+2xy+3xy²

       ^{=\(5x^2y+xy^2+2xy\)}

     N=2x²y+xy+xy²-4xy²-5xy

     =\(2x^2y-3xy^2-4xy\)

b) M-N=(\(5x^2y+xy^2+2xy\))-(\(2x^2y-3xy^2-4xy\))

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

           =\(3x^2y+4xy^2+6xy\)

M+N=\(5x^2y+xy^2+2xy\)\(+\)\(2x^2y-3xy^2-4xy\)

        =\(7x^2y-2xy^2-2xy\)

c) Ta có P(x)=0

\(\Rightarrow\)6-2x=0

\(\Rightarrow\)x=3

Vậy x=3 là nghiệm của đa thức P(x)

kcj ạ 

a) (3x2 – 2x2y) : x2 – (2xy2 + x2y) : (1/3 xy)

= (3x3 : x2) + (-2x2y : x2) - [(2x2y : 1/3 xy) +( x2y : 1/3 xy)]

= 3x – 2y – (6y + 3x) = 3x – 2y – 6y – 3x = -8y

A-B

=3x^2y^3-5x^3y^2-5xy+1-5x^3y^2+2x^2y^3+5xy-2

=5x^2y^3-10x^3y^2-1

5050 nha bài này tui hc trong lớp  4 ý :))

Ta có

4 x 2 y − 2 x y 2 + 1 3 x 2 y − x + 2 x 2 y + x y 2 − 1 3 x − 6 x 2 y = 4 x 2 y + 1 3 x 2 y + 2 x 2 y − 6 x 2 y + − 2 x y 2 + x y 2 + − 1 3 x − x = 1 3 x 2 y − x y 2 − 4 3 x

Chọn đáp án B

mn giúp e ik mn

\(a\text{)}x^2y+xy^2=xy\left(x+y\right)\)

\(b\text{)}x^2-2x+1=\left(x-1\right)^2\)

\(c\text{)}x^2-5x+4=\left(x-1\right)\left(x-4\right)\)

Bài 1: 

a: \(=6x^3-10x^2\)

b: \(=6x+5\)

Bài 1:

\(a,2x^2y\left(2x^2y^2-xy^2\right)\\ =2x^2x^2y^2y-2x^2x.y^2.y=2x^4y^3-2x^3y^3\\ b,\left(x-1\right)\left(2x+3\right)\\ =x.2x+x.3-1.2x-1.3=2x^2+3x-2x-3\\ =2x^2+x-3\\ c,\left(20x^3y^4+10x^2y^3-5xy\right):5xy\\ =20x^3y^4:5xy+10x^2y^3:5xy-5xy:5xy\\ =\left(20:5\right).\left(x^3:x\right).\left(y^4:y\right)+\left(10:5\right).\left(x^2:x\right).\left(y^3:y\right)-\left(5:5\right).\left(x:x\right).\left(y:y\right)\\ =4x^2y^3+2xy^2-1\\ d,\left(y-3x\right)^2-\left(y^2-6xy\right)\\ =\left[y^2-2.y.3x+\left(3x\right)^2\right]-\left(y^2-6xy\right)\\ =y^2-6xy+9x^2-y^2+6xy =9x^2\)

Bài 2:

\(a,4xy+4xz=4x\left(y+z\right)\\ b,x^2-y^2+9-6x\\ =\left(x^2-6x+9\right)-y^2\\ =\left(x-3\right)^2-y^2\\ =\left(x-3-y\right)\left(x-3+y\right)\)

Bài 3:

\(a,\dfrac{3xy}{y+z}+\dfrac{3xz}{y+z}\\=\dfrac{3xy+3xz}{y+z}\\ =\dfrac{3x\left(y+z\right)}{\left(y+z\right)}=3x\left(Với:y\ne-z\right)\\ b,\dfrac{x}{x+2}-\dfrac{x}{x-2}\\ =\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x^2-2x}{\left(x+2\right)\left(x-2\right)}=0\)

Bài 4:

a: \(A=\dfrac{3}{2}x^2+6x+3\)

b: \(B=5xy-\dfrac{2}{3}xy+\dfrac{1}{2}x^2y+2x^2y=\dfrac{5}{2}x^2y+\dfrac{13}{3}xy\)

a) \(2x^2+x-\dfrac{1}{2}x^2+5x+3\)\(\)

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

= \(\dfrac{3}{2}x^2+6x+3\)

Vậy A = \(\dfrac{3}{2}x^2+6x+3\)