b) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\dfrac{\left(2x+y\right)^2}{2x+y}=2x+y\)

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

\(a,VP=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ =\left(x+2y\right)\left[x^2-x.2y+\left(2y\right)^2\right]\\ =x^3+\left(2y\right)^3=x^3+8y^3=VT\left(đpcm\right)\\ b,VT=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)\\ =x^3-y^3-3xy\left(x-y\right)\\ =x^3-3x^2y+3xy^2-y^3\\ =\left(x-y\right)^3=VP\left(đpcm\right)\)

\(c,VT=\left(x-3y\right)\left(x^2+3xy+9y^2\right)-\left(3y+x\right)\left(9y^2-3xy+x^2\right)\\ =\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]-\left(x+3y\right).\left[x^2-x.3y+\left(3y\right)^2\right]\\ =x^3-27y^3-\left(x^3+27y^3\right)\\ =-54y^3=VP\left(đpcm\right)\)

Bài 6a) x^3 - 8

c) x^3 - 27y^3

d) = x^6 - 27

Bài 5a) 6859

b) 8120601

\(5,\\ a,19^3=\left(20-1\right)^3=8000-3\cdot400\cdot1+3\cdot20\cdot1-1\\ =8000-1200+60-1=6859\\ b,201^3=\left(200+1\right)^3=8000000+3\cdot40000\cdot1+3\cdot200\cdot1+1\\ =8000000+120000+600+1=8120601\)

\(=x^3+64\\ =x^3-27y^3\\ =x^6-\dfrac{1}{27}\)

\(\left(x+4\right)\left(x^2-4x+16\right)=x^3+64\)

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

\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)

a) = (x - 4y)(x + 1)

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

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

c) = x^2(x + 3) - 7x(x + 3) + 9(x + 3)

= (x + 3)(x^2 - 7x + 9)

a: \(x^2-4xy+x-4y\)

\(=x\left(x-4y\right)+\left(x-4y\right)\)

\(=\left(x-4y\right)\left(x+1\right)\)

b: \(x^2-6xy+9y^2-4\)

\(=\left(x-3y\right)^2-4\)

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

Chọn B.

Ta có x² + 9y² = 6xy  tương đương (x - 3y) ² = 0 hay x = 3y.

Khi đó 

a ) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)\)

\(=\left(2x+y\right)^2:\left(2x+y\right)\)

\(=2x+y\)

b ) \(\left(27x^3+1\right):\left(3x+1\right)\)

\(=\left(3x+1\right)\left(9x^2-3x+1\right):\left(3x+1\right)\)

\(=9x^2-3x+1\)

c ) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)\)

\(=\left(x-3y\right)^2:\left(3y-x\right)\)

\(=\left(3y-x\right)^2:\left(3y-x\right)\)

\(=3y-x\)

d ) \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)

\(=\left(2x-1\right)\left(4x^2+2x+1\right):\left(4x^2+2x+1\right)\)

\(=2x-1\)

:D

Đáp án A.

Ta có x² + 9y² = 6xy  <=> (x – 3y)² = 0 <=> x = 3y.

⇒ M = 1 + log 12   x + log 12   y 2 . log 12   6 y = log 12   12 + log 12   3 y 2 log 12   36 y 2

= log 12   36 y 2 log 12   36 y 2 = 1 .