Question

phân tích thành nhân tử :     \(x^5+x^4+1\) và  \(x^8+x+1\)

Answer 1

x⁸+x+1=x⁸-x⁵+x⁵-x²+x²+x+1

=x⁵.(x³-1)+x².(x³-1)+(x²+x+1)

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

=(x²+x+1)(x⁶-x⁵+x³-x²+1)

Answer 2

= x^5 - x^3 - x² - x^4 + x² + x + x^3 - x - 1 

= x²( x^3 - x - 1 ) - x( x^3 - x - 1 ) + ( x^3 - x - 1 ) 

= ( x² - x + 1)( x^3 - x - 1 )

Answer 3

Sai òi Phương Nam Võ! Nek

x^5+x^4+1= (x^5+x^4+X^3)-(x^3+x^2+x)+(x^2+x+1)

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

               =(x^2+x+1)(x^3-x+1)

Answer 4

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

Answer 5

x⁸ + x + 1

= x⁸  - x² + x²  + x + 1

= x²(x⁶-1) + x²+x+1

=x²(x³-1)(x³+1) + x²+ x + 1

=..................

.....

Answer 6

những bài này lop8 ma 

Answer 7

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

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

=(x²+x+1)(x⁶-x⁵+x³-x²+1)