x⁴y⁴ + 4

= x⁴y⁴ + 4x²y² + 4 - 4x²y²

= (x²y² + 2)² - (2xy)²

= (x²y² - 2xy + 2)(x²y² + 2xy + 2)

x⁴y⁴ + 64

= x⁴y⁴ + 16x²y² + 64 - 16x²y²

= (x²y² + 8)² - (4xy)²

= (x²y² - 4xy + 8)(x²y² + 4xy + 8)

x⁵ + x + 1

= x⁵ - x² + x² + x + 1

= x²(x³ - 1) + (x² + x + 1)

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

= (x² + x + 1)[x²(x - 1) + 1]

      \(x^5+x+1\)

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

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

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

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

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

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

k mk nha

bạn ơi bạn chưa bớt 2x^2 kìa

ngu dốt

x⁵-x⁴-1=x⁵-x³-x²-x⁴+x²+x+x³-x-1

=x².(x³-x-1)-x.(x³-x-1)+(x³-x-1)

=(x³-x-1)(x²-x+1)

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

k mk nha

mik chịu

Ta có : x⁵ + x + 1 

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

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

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

= (x² + x + 1)(x³ - x² + 1) . 

Ta có : x⁵ + x + 1 

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

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

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

= (x² + x + 1)(x³ - x² + 1) . 

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

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

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

= (x² + x + 1)(x³ - x² + 1) . 

x^4+4=x^4 + 4x^2 +4 - 4x^2=(x^2)^2+ 2.x^2.2+2^2 - (2x)^2 = (x^2+2)-(2x)^2 =(x^2+2-2x)(2^2+2-2x)

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

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

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

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

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

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

a) \(x^5+x+1=x^5+x^2-x^2+x+1\)

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

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

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

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

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

b) \(x^7+x^2+1=x^7+x^2-x+x+1\)

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

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

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

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

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

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

(Nếu đúng thì k cho mìk với nhé!)

a, \(x^5+x^4+1\)

\(\Leftrightarrow x^5+x^4-x^2+\frac{1}{4}-\frac{1}{4}+x^2\)

\(\Leftrightarrow x^5+\left(x^2-\frac{1}{2}\right)^2-\frac{1}{4}+x^2\)

\(\Leftrightarrow x^2\left(x^3+1\right)+\left(x^2-\frac{1}{2}\right)^2-\frac{1}{4}\)

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

ta có :x^5 +x^4 +1=x^5-x^2 +x^4 -x +x^2 +x +1=x^2(x^3-1) +x(x^3 -1)+x^2 +x +1=x^2(x-1)(x^2+x+1)+x(x-1)(x^2 +x+1) +x^2 +x +1=(x^2 +x +1)(x^3 -x^2 +x^2 -x +1)=(x^2 +x+1)(x^3-x+1)

ta có x^5 +x +1=x^5-x^2 +x^2+x+1=x^2(x^3-1) +x^2 +x+1=x^2(x-1)(x^2+x+1)+x^2+x+1=(x^2+x+1)(x^3-x^2+1)

a, 3x^2 + 13x + 10  

= 3x^2 + 3x + 10x + 10 

= 3x(x + 1) + 10(x + 1)

= (3x + 10)(x + 1)

b, x^2 - 10x + 21

= x^2 - 3x - 7x + 21

= x(x - 3) - 7(x - 3)

= (x - 7)(x - 3)

c, 6x^2 - 5x + 1

= 6x^2 - 3x - 2x + 1

= 3x(2x - 1) - (2x - 1)

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

Bạn đăng 1 lần nhiều bài như vậy làm người khác nản lắm đấy =) đơn giản bài rất dài mà mik cx ko chắc là bản thân mik có đc k hay ko nên phải nản vậy thôi :)

1a)\(3x^2+13x+10=3x^2+3x+10x+10\)

\(3x\left(x+1\right)+10\left(x+1\right)=\left(3x+10\right)\left(x+1\right)\)

b)\(x^2-10x+21=x^2-3x-7x+21\)

\(=x\left(x-3\right)-7\left(x-3\right)=\left(x-7\right)\left(x-3\right)\)

c)\(6x^2-5x+1=6x^2-3x-2x+1\)

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

1,

\(a,3x^2+13x+10=3x^2+3x+10x+10\)

\(=3x\left(x+1\right)+10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x+10\right)\)

\(b,x^2-10x+21=x^2-7x-3x+21\)

\(=x\left(x-7\right)-3\left(x-7\right)\)

\(=\left(x-7\right)\left(x-3\right)\)

\(c,6x^2-5x+1=6x^2-2x-3x+1\)

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

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

2,

\(a,x^8+4=\left(x^4\right)^2+4x^4+2^2-4x^4\)

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

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

\(b,x^4+18^2=\left(x^2\right)^2+36x^2+18^2-36x^2\)

\(=\left(x^2+18\right)-36x^2\)

\(=\left(x^2+18-6x\right)\left(x^2+18+6x\right)\)

\(c,x^4+3x^2+4=x^4+4x^2+4-x^2\)

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

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

Toán Học Team