a: Sửa đề: \(2A+\left(2x^2+y^2\right)=6x^2+5y^2-2x^2y^2\)

=>\(2A=6x^2+5y^2-2x^2y^2-2x^2-y^2\)

=>\(2A=4x^2+4y^2-2x^2y^2\)

=>\(A=2x^2+2y^2-x^2y^2\)

b: \(2A-\left(xy+3x^2-2y^2\right)=x^2-8y+xy\)

=>\(2A=x^2-8y+xy+xy+3x^2-2y^2\)

=>\(2A=4x^2+2xy-8y-2y^2\)

=>\(A=2x^2+xy-4y-y^2\)

c: Sửa đề: \(A+\left(3x^2y-2xy^2\right)=2x^2y+4xy^3\)

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

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

1.

\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)

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

\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)

\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)

\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)

\(=\left(x+y\right)^3\)

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

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

\(=\left(x-y\right)^3\)

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

2.

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

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

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

\(=x^3+y^3\)

\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)

\(=24xy+4x-6y-1-24xy-4x\)

\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)

\(=-6y-1\)

#Toru

Tất cả là đơn thức đồng dạng

Mik k hỉu lắm!

`Answer:`

Gọi đơn thức cần tìm là `n`

Thay vào ta được: `2x^2y + n = -4x^2y`

`<=>n=-4x^2y - 2x^2y`

`<=>n=-6x^2y`

Vậy đơn thức cần tìm là `-6x^2y`

`=>` Chọn C.

Đề là gì nhỉ

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

Mik giống ''Nguyễn Hoàng Minh'' nha.

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

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

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

\(=6x^4y^2:\dfrac{1}{4}x^4y^2=24\)

\(\dfrac{6x^4y^2}{\left(\dfrac{1}{2}x^2y\right)^2}=\dfrac{6x^4y^2}{\dfrac{1}{4}x^4y^2}=24\)