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Bài 1:
\(x^5+x+1\)
\(=x^5-x^4+x^2+x^4-x^3+x+x^3-x^2+1\)
\(=x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Bài 2:
\(\frac{2n^2-3n+1}{2n+1}=\frac{n\left(2n+1\right)-4n+1}{2n+1}=\frac{n\left(2n+1\right)}{2n+1}-\frac{4n+1}{2n+1}=n-\frac{4n+1}{2n+1}\in Z\)
\(\Rightarrow4n+1⋮2n+1\)
\(\Rightarrow\frac{4n+1}{2n+1}=\frac{2\left(2n+1\right)-1}{2n+1}=\frac{2\left(2n+1\right)}{2n+1}-\frac{1}{2n+1}=2-\frac{1}{2n+1}\in Z\)
\(\Rightarrow1⋮2n+1\)
\(\Rightarrow2n+1\inƯ\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow2n\in\left\{0;-2\right\}\)
\(\Rightarrow n\in\left\{0;-1\right\}\)
\(3m^2-2m-1\)
\(=3m^2-3m+m-1\)
\(=3m\left(m-1\right)+\left(m-1\right)\)
\(=\left(m-1\right)\left(3m+1\right)\)
\(3m^2-2m-1\)
\(=3m^2+m-3m-1\)
\(=\left(3m^2+m\right)-\left(3m+1\right)\)
\(=m\left(3m+1\right)-\left(3m+1\right)\)
\(=\left(m-1\right)\left(3m+1\right)\)
<=>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
mình chỉ phân tích được đa thức này thôi!
\(x^4+x^2+1\)
\(=x^4+2x^2-x^2+1\)
\(=\left(x^4+2x^2+1\right)-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
\(x^4+x^2+1\)
\(=x^4+2x^2+1+x^2-2x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+1-x\right).\left(x^2+1+x\right)\)
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 )
\(x^7+x^2+1\)
\(=x^7+x^6+x^5+x^4+x^3+x^2+x+1\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
a) \(x^7+x^2+1=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
b) \(x^7+x^5+1=\left(x^7+x^6+x^5\right)-\left(x^6-1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x^2+x+1\right)\left[x^5-\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
a(x2 + 1) - x(a2 + 1)
= ax2 + a - a2x - x
= (ax2 - a2x) + (a - x)
= -ax(a - x) + (a - x)
= (a - x)(-ax + 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 ^ _ ^
\(x^4+x^3+x^2-1\)
\(=x^2\left(x^2-1\right)+x^2-1\)
\(=\left(x^2+1\right)\left(x^2-1\right)\)