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)M= 3,5x²y-2xy+1,5x²y+2xy+3xy²

M= (3,5x²y+1,5x²y)+(-2xy+2xy)+3xy²

M=5x²y+3xy²

N= 2x²y+3,2xy+xy²-4xy²-1,2xy

N= (xy²-4xy²)+(3,2xy-1,2xy)+2x²y

N=-3xy²+2xy+2x²y

b) ta có M=5x²y+3xy² (đã thu gọn)

N=-3xy²+2x²y+2xy (đã thu gọn)

=> M-N=(5x²y+3xy²)+(-3xy²+2x²y+2xy)

M-N=5x²y+3xy²-3xy²+2x²y+2xy

M-N=(5x²y+2x²y)+(3xy²-3xy²⁾+2xy

M-N=7x²y+2xy

Hy vọng là đúng ạ!!!

a: A = -2xy + 3/2xy^2 + 1/2xy^2 + xy = -2xy + 2xy^2 + xy = 2xy^2 - xy

b: B = xy^2z + 2xy^2z - xyz - 3xy^2z + xy^2z = 3xy^2z - xyz

c: C = 4x^2y^3 + x^4 - 2x^2 + 6x^4 - x^2y^3 = 7x^4 + 3x^2y^3 - 2x^2

d: D = 3/4xy^2 - 2xy - 1/2xy^2 + 3xy = 5/4xy^2 + xy

e: E = 2x^2 - 3y^3 - z^4 - 4x^2 + 2y^3 + 3z^4 = -2x^2 - y^3 + 2z^4

f: F = 3xy^2z + xy^2z - xyz + 2xy^2z - 3xyz = 6xy^2z - 2xyz

a: A=-2xy+3/2xy^2+1/2xy^2+xy

=-2xy+xy+3/2xy^2+1/2xy^2

=2xy^2-xy

b: \(B=xy^2z+2xy^2z-xyz-3xy^2z+xy^2z\)

\(=xy^2z\left(1+2-3+1\right)-xyz=xy^2z-xyz\)

c: \(=4x^2y^3-x^2y^3+x^4+6x^4-2x^2\)

\(=7x^4-x^2+3x^2y^3\)

d: \(=\dfrac{3}{4}xy^2-\dfrac{1}{2}xy^2+3xy-2xy\)

=1/4xy^2+xy

e: \(=2x^2-4x^2-3y^3+2y^3+3z^4-z^4\)

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

f: \(=xy^2z+3xy^2z+2xy^2z-xyz-3xyz\)

\(=6xy^2z-4xyz\)

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

Ta có: \(A+B+C=0\)

\(\Leftrightarrow3x^2y+5xy^2-2xy+1+2x^2y-7xy^2+6xy-8-5x^2y+4xy^2-4xy+12=0\)

\(\Leftrightarrow2xy^2+5=0\)

\(\Leftrightarrow2x\cdot\left(-2\right)^2+5=0\)

\(\Leftrightarrow8x+5=0\)

\(\Leftrightarrow8x=-5\)

hay \(x=-\dfrac{5}{8}\)

Vậy: \(x=-\dfrac{5}{8}\)

a: \(=x^2\left(2x+3\right)+\left(2x+3\right)\)

\(=\left(2x+3\right)\left(x^2+1\right)\)

b: \(=\left(x-4\right)\left(x+3\right)\)

e: =(x+3)(x-2)

a) \(=x^2\left(2x+3\right)+\left(2x+3\right)=\left(2x+3\right)\left(x^2+1\right)\)

b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)

c) \(=\left(2x\right)^2-\left(x^2+1\right)^2=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)

d) \(=4xy\left(y-3x+2\right)\)

e) \(=x\left(x-2\right)+3\left(x-2\right)=\left(x-2\right)\left(x+3\right)\)

f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)

g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)

h) \(=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)

i) \(=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)^2\left(x+3\right)\)

M-N-P=4x³-2x²y+xy+1-3x²y-2xy+5-4x³+5x²y-3xy-1

           =-4xy+5

p-n-m=4x³-5x²y+3x²y+1-3x²y-2xy+5-4x³+2x²y-xy-1

          =-6x²y+5

\(a)P=3,5.x^2y-3.x.y^2+1,5.x^2.y+2.x.y+3.x.y^2\)

\(P=5.x^2.y+2.x.y\)

\(b)\text{Thay x=1;y=2 vào biểu thức P,ta được:}\)

\(5.1^2.2+2.1.2\)

\(=5.1.2+2.1.2\)

\(=10+4=14\)

\(\text{Vậy giá trị của biểu thức P tại x=1;y=2 là:14}\)

a.\(P=3,5x^2y-3xy^2+1,5x^2y+2xy+3xy^2\)

\(P=5x^2y+2xy\)

b. Thế x=1; y=2 vào P, ta được:

\(5.1^2.2+2.1.2=10+4=14\)

Ta có :