\(8x^2y-8xy-2yz^2+2y\)

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

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

\(8x^3-\frac{1}{8}\)

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

\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

tích hộ

Ta có

8 x 3   +   12 x 2 y   +   6 x y 2   +   y 3     =   ( 2 x ) 3   +   3 . ( 2 x ) 2 y   +   3 . 2 x . y 2   +   y 3   =   ( 2 x   +   y ) 3

Đáp án cần chọn là: B

a) Áp dụng HĐT 5 thu được ( 2 a   -   3 b ) 3 .

b) Ta có  8 x 3   +   12 x 2 y   +   6 xy 2   +   y 3  = ( 2 x   +   y ) 3 .

Áp dụng HĐT 7 với A = 2x + y; B = z

( 2 x   +   y ) 3 - z 3 = (2x + y - z)(4 x 2   +   y 2   +   z 2  + 4xy + 2xz + zy).

a) \(4x^3y^2-8x^2y+12xy^2=4xy\left(x^2y-2x+3y\right)\)

b) \(3x^2-6xy-5x+10y=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)

c) \(x^2-49+4y^2-4xy=\left(x-2y\right)^2-49=\left(x-2y-7\right)\left(x-2y+7\right)\)

d) \(x^2-6x-16=\left(x^2-6x+9\right)-25=\left(x-3\right)^2-25=\left(x-3-5\right)\left(x-3+5\right)=\left(x-8\right)\left(x+2\right)\)

a) 

b) 

c) 

d) 

a) \(4x^3y^2-8x^2y+12xy^2=4xy.x^2y-4xy.2x+4xy.3y=4xy\left(x^2y-2x+3y\right)\)

b) \(3x^2-6xy-5x+10y=\left(3x^2-6xy\right)-\left(5x-10y\right)=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)

c) \(x^2-49+4y^2-4xy=\left(x^2-4xy+4y^2\right)-49=\left(x-2y\right)^2-7^2=\left(x-2y-7\right)\left(x-2y+7\right)\)

d) \(x^2-6x-16=\left(x^2-8x\right)+\left(2x-16\right)=x\left(x-8\right)+2\left(x-8\right)=\left(x-8\right)\left(x+2\right)\)

1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2, \(x^2-10x+25=\left(x-5\right)^2\) 

3, \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

4, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)

1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2) \(x^2-10x+25=\left(x-5\right)^2\)

3) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2x}{3}+\dfrac{1}{4}\right)\)

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

1: \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2: \(x^2-10x+25=\left(x-5\right)^2\)

3: \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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

a) Ta có: \(1+6x-6x^2-x^3\)

\(=\left(1-x\right)\left(x^2+x+1\right)+6x\left(1-x\right)\)

\(=\left(1-x\right)\left(x^2+7x+1\right)\)

b:Ta có: \(x^3-4x^2+8x-8\)

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

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

c: Ta có: \(x^3+2x^2+2x+1\)

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

\(=\left(x+1\right)\left(x^2+x+1\right)\)

d: Ta có: \(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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

e: \(x^2+6x+9-y^2\)

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

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

f: \(x^2-2x+7x-14\)

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

=(x-2)(x+7)

h: \(5x^2-10xy+5y^2-20\)

\(=5\left(x^2-2xy+y^2-4\right)\)

\(=5\left(x-y-2\right)\left(x-y+2\right)\)

a: \(3x^4-6x^3+2x^2=x^2\left(3x^2-6x+2\right)\)

b: \(x^3y+12x^2y+36xy=xy\left(x^2+12x+36\right)=xy\left(x+6\right)^2\)

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

d: \(x^2y^2-2xy^2+y^2=y^2\left(x-1\right)^2\)