Đặt (a + b)³ = a³ + 3a²b + 3ab² + b³ = 8x³ + 12x² + ... + ...

                                                          = (2x)³ + 3.(2x)².1 + ... + ...

=> a = 2x ; b = 1 => 8x³ + 12x² + 6x + 1 = (2x + 1)

a, 15x³y⁵z : 5x²y³ = 3xy²z.

b, 12x⁴y² : ( - 9xy² ) = \(\frac{3}{4}x^3\).

c, ( 30x⁴y³ - 25x²y³ - 3x⁴y⁴ ) : 5x²y³ = \(6x^2-5-\frac{3}{5}x^2y.\)

d, ( 4x⁴ - 8x²y² + 12x⁵y ) : ( - 4x² ) = -x² + 2y² - 3x³y.

a: \(=\dfrac{15}{5}\cdot\dfrac{x^3}{x^2}\cdot\dfrac{y^5}{y^3}\cdot z=3xy^2z\)

b: \(=-\dfrac{4}{3}x^3\)

c: \(=\dfrac{30x^4y^3}{5x^2y^3}-\dfrac{25x^2y^3}{5x^2y^3}-\dfrac{3x^4y^4}{5x^2y^3}\)

\(=6x^2-5-\dfrac{3}{5}x^2y\)

d: \(=\dfrac{4x^4}{-4x^2}+\dfrac{8x^2y^2}{4x^2}-\dfrac{12x^5y}{4x^2}\)

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

a: =-4xyz^2

b: =-9x^2y

c: =16x^2y^2

d: =1/6x^2y^3

e: =13/6x^3y^2

f: =7/12x^4y

a) -xyz² - 3xz.yz

= -xyz² - 3xyz²

= -4xyz²

b) -8x²y - x.(xy)

= -8x²y - x²y

= -9x²y

c) 4xy².x - (-12x²y²)

= 4x²y² + 12x²y²

= 16x²y²

d) 1/2 x²y³ - 1/3 x²y.y²

= 1/2 x²y³ - 1/3 x²y³

= 1/6 x²y³

e) 3xy(x²y) - 5/6 x³y²

= 3x³y² - 5/6 x³y²

= 13/6 x³y²

f) 3/4 x⁴y - 1/6 xy.x³

= 3/4 x⁴y - 1/6 x⁴y

= 7/12 x⁴y

\(\left(2x-y\right)^3=\left(2x\right)^3-3\left(2x\right)^2y+3\cdot2x\cdot y^2-y^3\\ =8x^3-12x^2y+6xy^2-y^3\)

Chọn \(12x^2y\)

\(a,\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=8x^3-27y^3\)

\(b,\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)

\(c,64-48x+12x^2-x^3=\left(4-x\right)^3\)

\(d,x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)

\(e,8x^3-60x^2y+150xy^2-125y^3=\left(2x-5y\right)^3\)

trả lời muộn quá đấy uyên ạ -_-

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