271​x3−32​x2+4x−8

 x  -  x  +  4x  -  8

=  4x  -  8

=  4( x  -  8)

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

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

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

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

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

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

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

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

192=(20−1)2=202−2.20.1+12=400−40+1=361

a)

19^2=(20−1)^2=20^2−2.20.1+1^2=400−40+1=361

28^2=(30−2)^2=30^2−2.30.2+2^2=900−120+4=784

81^2=(80+1)^2=80^2+2.80.1+1^2=6400+160+1=6561

91^2=(90+1)^2=90^2+2.90.1+1^2=8100+180+1=8281

c) (xy^2+1)^2

d) (1/3-y^4)^2

e) (1/2a-2b^2)^2

f) (5-x)^2

c) 2xy² + x² y⁴ + 1

= (xy²)² + 2xy² .1 + 1²

=(xy² +1)²

d) 1/9 - 2/3 y⁴ + y⁸

= (1/3)²  - 2 . 1/3 y⁴ + (y⁴)²

=(1/3 - y⁴)²

e) 1/4 a² - 2ab²  + 4b⁴

= (1/2 a)² - 2 1/2 a . 2b  + (2b²)²

=(1/2 a  - 2b²)²

f) 25 - 10x + x²

= x² + 2 . 5x + 5²

= (x+5)²

x² + x + 1/4 = ( x + 1/2 )²

Trả lời:

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

...

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

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

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

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

=[(x^4+x)(x-1)+1](x^2+x+1)

=(x^5-x^4+x^2-x)(x^2+x+1)

Trả lời:

1, x⁷ + x² + 1 

= x⁷ + x² + 1 + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x 

= ( x⁷ + x⁶ + x⁵ ) - ( x⁶ + x⁵ + x⁴ ) + ( x⁴ + x³ + x² ) - ( x³ + x² + x ) + ( x² + x + 1 )

= x⁵ ( x² + x + 1 ) - x⁴ ( x² + x + 1 ) + x² ( x² + x + 1 ) - x ( x² + x + 1 ) + ( x² + x + 1 )

= ( x² + x + 1 )( x⁵ - x⁴ + x² - x + 1 )

b, x⁸ + x⁷ + 1 

= x⁸ + x⁷ + 1 + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x 

= ( x⁸ + x⁷ + x⁶ ) - ( x⁶ + x⁵ + x⁴ ) + ( x⁵ + x⁴ + x³ ) - ( x³ + x² + x ) + ( x² + x + 1 ) 

= x⁶ ( x² + x + 1 ) - x⁴ ( x² + x + 1 ) + x³ ( x² + x + 1 ) - x ( x² + x + 1 ) + ( x² + x + 1 )

= ( x² + x + 1 )( x⁶ - x⁴ + x³ - x + 1 )

=(x+y)²+2(x+y)+1-16

=(x+y+1)²-4²

=(x+y+5)(x+y-3)

Sửa đề: \(x^2+2xy+y^2+2x+2y-15\)

\(=\left(x+y\right)^2+2\left(x+y\right)+1-16\)

Đặt \(x+y=t\)

\(\Rightarrow t^2+2t+1-16\)

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

\(=\left(t+1-4\right)\left(t+1+4\right)\)

\(=\left(t-3\right)\left(t+5\right)\)

\(=\left(x+y-3\right)\left(x+y+5\right)\)

\(1,\)

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

Đặt: \(\left(x^2-x+2\right)^2=n\)

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

\(=n^2-3x^2n+2x^4\)

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

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

Thay \(\left(x^2-x+2\right)^2\)ta có:

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

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

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

\(2,\)

\(3\left(-x^2+2x+3\right)^4-26x^2\left(-x^2+2x+3\right)^2-9x^4\)

\(=3\left(-x^2+2x+3\right)^4+x^2\left(-x^2+2x+3\right)^2-27x^2\left(-x^2+2x+3\right)^2-9x^4\)

\(=\left(-x^2+2x+3\right)^2[3\left(-x^2+2x+3\right)^2+x^2]-9x^2[\left(-x^2+2x+3\right)^2+x^2]\)

\(=[3\left(-x^2+2x+3\right)^2+x^2][\left(-x^2+2x+3\right)^2-9x^2]\)

\(x^2+2xy+y^2+2x+2y-15\)

\(=\left(x+y\right)^2+2\left(x+y\right)+1-16\)

\(=\left(x+y+1\right)^2-4^2=\left(x+y-3\right)\left(x+y+5\right)\)

\(x^2+2xy+y^2+2x+2y-15\)

\(=x^2+2xy+y^2+2x+2y+1-16\)

\(=\left(x+y\right)^2+2\left(x+y\right)+1-16\)

Đặt \(x+y=t\)

\(\Rightarrow t^2+2t+1-16\)

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

\(=\left(t+1-4\right)\left(t+1+4\right)\)

\(=\left(t-3\right)\left(t+5\right)\)

\(=\left(x+y-3\right)\left(x+y+5\right)\)

phân tích đa thức thành nhân tử 4x^3+6x^2+4x+1

= 4x^3+6x^2+4x+1

= (2x+1)(2x^2+2x+1)

nha bạn chúc bạn học tốt nha 

\(4x^3+6x^2+4x+1\)

\(=4x^3+4x^2+2x^2+2x+2x+1\)

\(=\left(4x^3+4x^2+2x\right)+\left(2x^2+2x+1\right)\)

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

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