a) 2x(y-z)-6y(z-y)

=2x(y-z)+6y(y-z)

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

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

=(x-y)(x+y)+4(x-y)

=(x-y)(x+y+4)

P/s tham khảo nha

a)  = 2x(y-z)-6y(y-z)

= 2(y-z)(x-3y)

b)  = 

\(=\dfrac{2\left(x-2y+z\right)^3+4\left(x-2y+z\right)^2}{2\left(x-2y+z\right)}=\left(x-2y+z\right)^2+2\left(x-2y+z\right)\)

Anh chỉ em pp học Toán giỏi như anh được ko ạ  rất thích môn Toán nên cũng rất hâm mộ anh

4x(x+y)(x+y+z)(x+z)+y²z²=4(x²+xy+xz)(x²+xy+xz+yz)+y²z²=4(x²+xy+xz)²+4yz(x²+xy+xz)+y²z²=(2(x²+xy+xz)+yz)²=(2x²+2xy+2xz+yz)

Câu hỏi của Lee Min Ho - Toán lớp 8 - Học toán với OnlineMath

Câu hỏi của Lee Min Ho - Toán lớp 8 - Học toán với OnlineMath

x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz

=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz

=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz

=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3

=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]

=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)

=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]

=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]

=(x+y+z)(x-y-z)(z-x-y)

x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz

=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz

=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz

=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3

=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]

=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)

=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]

=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]

=(x+y+z)(x-y-z)(z-x-y)

b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)

\(2x^2+2y^2-x^2z+z-y^2z-2\)

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

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

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

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