tớ thử nha
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)
Ta có:\(x^5+x+1\)
\(=\left(x^5-x^2\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
\(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^2+x+1\right)\)
\(\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)
x5+x-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)