\(=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)

x(x-4)+5(x-4)

(x+5)(x-4)

\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left[\left(x-1\right)\left(x-7\right)\right].\left[\left(x-3\right)\left(x-5\right)\right]-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x+11=t\) \(\Rightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=\left(t-4\right)\left(t+4\right)-20=t^2-16-20=t^2-36=\left(t-6\right)\left(t+6\right)\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left(x^2-8x+11-6\right)\left(x^2-8x+11+6\right)=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)

a) \(x^7+x^2+1\)

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

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

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

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

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

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

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

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

b) \(x^7+x^5+1\)

\(=x^7+x^6+x^5-x^6+1\)

\(=\left(x^7+x^6+x^5\right)-\left(x^6-1\right)\)

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

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

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

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

\(=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)

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

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

\(x^7+x^5+1\)

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

k cho mình nha

\(\text{Đa thức bậc 7 = đa thức bậc 2 + đa thức bậc 5 }\)

\(\text{Chia theo sơ đồ Hoocner sẽ nhanh hơn}\)

\(\text{OoO cô bé tinh nghịch OoO làm đúng rồi nhé}\)

x^5+x+1

=x(x^4+1)+1 

=(x^2+x+1)(x^3-x^2+1) 

Ta có : x⁵ + x + 1 

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

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

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

= (x + 1)(x⁵ - x⁴ + x³ - x² + 1)

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

hc tốt nha !!!!!!!!!

KHÔNG BÍT

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

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