Câu bc mình ghi nhầm nên dừng làm

kết bạn với mình đi

hihihihihihhihihhihihihhihihihhihihhihi

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

\(P=-3x^2y^2+6xy^3-x^4+x^2y^2+2x^2y^2-6xy^3\)

\(P=-x^4\)

Thay x = -2 vào P, ta có:

\(P=-\left(-2\right)^4=-16\)

Ta có: \(P=-3xy\left(xy-2y^2\right)-x^2\left(x^2-y^2\right)+2y^2\left(x^2-3xy\right)\)

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

\(=-x^4\)

\(=-16\)

trả lời hộ với mai thi rồi

a: \(A=2\left(x+y\right)+3xy\left(x+y\right)+5x^2y^2\left(x+y\right)=0\)

b: \(B=3xy\left(x+y\right)+2x^2y\left(x+y\right)=0\)

\(\Rightarrow\)A=2(x+y)+3xy(x+y)+5x²y²(x+y)

Thay x+y=0 vào A

\(\Rightarrow\)A=0

\(5x\left(3x^2y-2xy^2+1\right)-3xy\left(5x^2-3xy\right)+x^2y^2-10=0\)

\(\Leftrightarrow15x^3y-10x^2y^2+5x-15x^3y+9x^2y^2+x^2y^2-10=0\)

\(\Leftrightarrow5x=10\Leftrightarrow x=2\)

a: =xy(1/3+4-2)=7/3xy

b: =xy^2(-1+3/2+4/3)=(1/3+3/2)xy^2=11/6xy^2

c: =4x^2y^2+2/3x^2y^2-4/3x^2y=-4/3x^2y+14/3x^2y^2

d: =3x^2y^2z+4x^2y^2z-8x^2y^2z=-x^2y^2z

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

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

=\(\left(\dfrac{-1}{10}\right)x^2y+\dfrac{4}{3}xy^2-6xy\)

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

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

\(=x\left(x-2y\right)-y\left(x-2y\right)+\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-y+1\right)\)