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\(x^4+x^2+1\)
\(=x^4+2x^2+1+x^2-2x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+1-x\right).\left(x^2+1+x\right)\)
Vì phương trình x4+x2+1=0 vô nghiệm nên không thể phân tích thành nhân tử
Ta có : x4 + x2 + 1
= x4 + x2 + x2 + 1 - x2
= (x2 + 1)2 - x2
= (x2 + 1 - x)(x2 + 1 + x)
x4 + x2 + 1
= x4 + 2x2 + 1 - x2
= ( x2 + 1 )2 - x2
= ( x2 - x + 1 )( x2 + x + 1 )
\(x^4+x^3+x^2-1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1\right)\left(x^3+\left(x-1\right)\right)\)
Ủng hộ nha ^ _ ^
\(x^4+x^3+x^2-1\)
\(=x^2\left(x^2-1\right)+x^2-1\)
\(=\left(x^2+1\right)\left(x^2-1\right)\)
Ta có : \(x^4+x^3+2x^2+x+1\)
\(=x^4+x^3+x^2+x^2+x+1\)
\(=x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+1\right)\)
\(x^4+x^2+1\)
\(=x^4+2x^2+1-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+1+x^2\right)\left(x^2+1-x^2\right)\)
\(=2x^2+1\)
\(A=x^4+x^2+1\)
\(=x^4-x+x^2+x+1\)
\(=x\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(x^5+x^4+2\)
\(=x^5+x^4+x^2-x^2+1+1\)
\(=\left(x^5-x^2\right)+\left(x^4+x^2+1\right)\)
\(=\left(x^5-x^2\right)+\left(x^4+2x^2-x^2+1\right)+1\)
\(=x^2\left(x^3-1\right)+\left(x^4+2x^2-x^2+1\right)+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(\left(x^2+1\right)^2-x^2\right)+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+1+x\right)\cdot\left(x^2+1-x\right)+1\)
\(=\left(x^3-x^2\right)\left(x^2+x+1\right)+\left(x^2+1+x\right)\cdot\left(x^2+1-x\right)+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+x^2+1-x\right)+1\)
\(=\left(x^2+x+1\right)\left(x^3+1-x\right)+1\)
<=>x4-x+x2 +x+1= x (x-1) (x2+x+1) + (x2+x+1) = (x2+x+1)(x2-x+1)
chắc có lẽ đúng đó