\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)

\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)

\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)

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

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

a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)

b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)

\(=y^2-2x^2y^3\)

c: \(=6x-y+2x^2+3y-2x^2+x\)

\(=7x+2y\)

d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)

a, mình nghĩ đề là cm đẳng thức nhé 

\(VT=\left(5x^4-3x^3+x^2\right):3x^2=\frac{5x^4}{3x^2}-\frac{3x^3}{3x^2}+\frac{x^2}{3x^2}=\frac{5}{3}x^2-x+\frac{1}{3}=VP\)

Vậy ta có đpcm 

b, \(VT=\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)=\frac{5xy^2}{-xy}+\frac{9xy}{-xy}-\frac{x^2y^2}{-xy}\)

\(=-5y-9+xy=VP\)

Vậy ta có đpcm 

c, \(VT=\left(x^3y^3-x^2y^3-x^3y^2\right):x^2y^2=\frac{x^3y^3}{x^2y^2}-\frac{x^2y^3}{x^2y^2}-\frac{x^3y^2}{x^2y^2}=xy-y-x=VP\)

Vậy ta có đpcm 

đăng lên làm j z

a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)

b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)

c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)

\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)

\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)

\(=y^2-2x^2y^3\)

\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)

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

\(=2x^2+7x+2y-2\)

\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)

\(=x-y+4y^2-6xy+10x^2\)

Oa, giờ mới biết bác cũng ở box Toán :))

a) (x²y² – xy + 2y)(x – 2y)

= x²y². X + x²y²(-2y) + (xy) . x + (-xy)(-2y) + 2y . x + 2y(-2y)

= x³y² – 2x²y³- x²y + xy² + 2xy – 4y²

b) (x² – xy + y²)(x + y) = x² . x + x². y + (-xy) . x + (-xy) . y + y² . x + y². y

= x³ + x². y - x². y - xy² + xy² + y³

= x³ - y³

a) (x²y² – xy + 2y)(x – 2y)

= x²y². X + x²y²(-2y) + (xy) . x + (-xy)(-2y) + 2y . x + 2y(-2y)

= x³y² – 2x²y³- x²y + xy² + 2xy – 4y²

b) (x² – xy + y²)(x + y) = x² . x + x². y + (-xy) . x + (-xy) . y + y² . x + y². y

= x³ + x². y - x². y - xy² + xy² + y³

= x³ - y³

a) biết chết liền

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

Bài giải:

a) 5x²y⁴ : 10x²y = $\frac{5}{10}$x^{2 – 2}. y^{4 – 1} = $\frac{1}{2}$y³

b) $\frac{3}{4}$x³y³ : (- $\frac{1}{2}$x²y²) = $\frac{3}{4}$ . (-2) . x^{3 – 2} . y^{3 – 2} = -$\frac{3}{2}$xy

c) (-xy)¹⁰ : (-xy)⁵ = (-xy)^{10 – 5} = (-xy)⁵ = -x⁵y⁵.