Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

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

=(x+y)(x-y-1)

b,   x²-2xy+y²-z²

=(x-y)²-z²

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

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

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

=4x²-y²+8y-16

=4x²-(y² - 8y+16)

=4x²-(y-4)²

=(2x+y-4)(2x-y+4)

   4x²-y²+8(y-2)

=4x²-y²+8.y-8.2

=4x²-y²+8y-16

Bạn trình bầy rõ ràng chút đi. Mk chẳng hiểu gì cả.

\(-25x^6-y^8+10x^3y^4\\ =-\left(25x^6-10x^3y^4+y^8\right)\\ =-\left[\left(5x^3\right)^2-2\cdot5x^3\cdot y^4+\left(y^4\right)^2\right]\\ \\ =\left(5x^3-y^4\right)^2\)

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

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

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

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

HỌC TỐT NHA!

ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)

- Hình như đề của u sai hay sao á :)))

Giải:

a) \(3x^2y-6xy^2\)

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

Vậy ...

b) \(\left(2x-a\right)x^2-\left(2x-a\right)y\)

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

\(=\left(2x-a\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)

Vậy ...

c) \(25a^2-c^2\)

\(=\left(5a-c\right)\left(5a+c\right)\)

Vậy ...

d) \(4-36x+81x^2\)

\(=2^2-2.2.9x+\left(9x\right)^2\)

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

Vậy ...

e) \(\left(x+7\right)2-\left(2x-9\right)2\)

\(=2\left[\left(x+7\right)-\left(2x-9\right)\right]\)

\(=2\left(x+7-2x+9\right)\)

\(=2\left(16-x\right)\)

Vậy ...

f) \(x^2-6x+8\)

\(=x^2-6x+9-1\)

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

\(=\left(x-4\right)\left(x-2\right)\)

Vậy ...

\(4x^2-y^2+8\left(y-2\right)=4x^2-y^2+8y-16=4x^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)

y² hay y ạ???