mình ko biết làm k bạn với mình nhé

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

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

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

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

b) \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

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

a) x⁶+x²y⁵+xy⁶+x²y⁵-xy⁶

= x⁶+(x²y⁵+x²y⁵)+(xy⁶-xy⁶)

= x⁶+2x²y⁵

b) \(\dfrac{1}{2}\)x²y³-x²y³+3x²y²z²-z⁴-3x²y²z²

= (\(\dfrac{1}{2}\)x²y³-x²y³)+(3x²y²z²-3x²y²z²)-z⁴

= -\(\dfrac{1}{2}\)x²y³-z⁴

0,2:x=1,03+3,97

a: A=-2xy+xy+xy^2=-xy+xy^2

Bậc là 3

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

Bậc là 4

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

Bậc là 5

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

bậc là 3

e: \(E=2x^2-4x^2+3z^4-z^4-3y^3+2y^3\)

=-2x^2+2z^4-y^3

Bậc là 4

f: \(=3xy^2z+xy^2z+2xy^2z-4xyz=6xy^2z-4xyz\)

Bậc là 4

1/ P(x)= x^4 + x^3 +x + 1

          = x^3(x+1)+(x+1) *1

          = (x+1)(x^3+1)

     Nghiệm P(x)khi P(x)=0

hay (x+1)(x^3+1)=0

suy ra x+1=0 do đó x=-1

và x^3+1=0 suy ra x^3=-1 nên x=-1

Vậy P(x) có 1 nghiệm là x=-1