Question

[nguyễn hữu kim]

các bạn giúp mình bài 4 nhé cảm ơn

Answer 1

[HT.Phong (9A5)]

Bài 3:

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

\(A=2xy^2\cdot x^3-2xy^2\cdot2xy+2xy^2\cdot2y^2\)

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

b) \(B=\left(x^2+y^2-z^2\right)\left(x^2+y^2+z^2\right)\)

\(B=x^2\cdot x^2+x^2\cdot y^2+x^2\cdot z^2+x^2\cdot y^2+y^2\cdot y^2+y^2\cdot z^2-x^2\cdot z^2-y^2\cdot z^2-z^2\cdot z^2\)

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

\(B=x^4+\left(x^2y^2+x^2y^2\right)+\left(x^2z^2-x^2z^2\right)+y^4+\left(y^2z^2-y^2z^2\right)-z^4\)

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

c) \(C=-\dfrac{1}{4}xy\left(4x^2y^2-x^2y-\dfrac{4}{5}\right)\)

\(C=-\dfrac{1}{4}xy\cdot4x^2y^2+\dfrac{1}{4}xy\cdot x^2y+\dfrac{1}{4}xy\cdot\dfrac{4}{5}\)

\(C=-x^3y^3+\dfrac{1}{4}x^3y^2+\dfrac{1}{5}xy\)

d) \(D=\left(x-y\right)^4\)

\(D=\left[\left(x-y\right)^2\right]^2\)

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

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

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

\(D=x^4+6x^2y^2+y^4\)

Answer 2

[Minh Lệ]

4/

a/ \(A=\dfrac{7y^5z^2-14y^3z^4+2,1y^4z^3}{-7y^3z^2}=\dfrac{7y^5z^2}{-7y^3z^2}+\dfrac{-14y^3z^4}{-7y^3z^2}+\dfrac{2,1y^4z^3}{-7y^3z^2}=-y^2+2z^2-0,3yz\)

b/ \(A=\dfrac{9x^3y+3xy^3-6x^2y^2}{-3xy}=\dfrac{9x^3y}{-3xy}+\dfrac{3xy^3}{-3xy}+\dfrac{-6x^2y^2}{-3xy}=-3x^2-y^2+2xy\)