Phân tích đa thức thành nhân tử:
1) x^2+2014x+2013
2)x(x+2)(x^2+2x+2)+1
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a: \(=x^2\left(x-y\right)+2014\left(x-y\right)=\left(x-y\right)\left(x^2+2014\right)\)
\(\left(x-1\right)^2-2\left(x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left(x-1-2x-1\right)^2=\left(-x-2\right)^2=\left(x+2\right)^2\)
\(\left(x^2-x+2\right)\left(x-1\right)-x^2\left(x-1\right)^2+\left(2x+1\right)\left(x-1\right)^3\)
\(=\left(x-1\right)\left[x^2-x+2-x^2\left(x-1\right)+\left(2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2-x+2-x^3+x^2+2x^3-4x^2+2x+x^2-2x+1\right)\)
\(=\left(x-1\right)\left(x^3-x^2-x+3\right)\)
Ta có \(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)\)
\(=1-4x^2-x\left(x^2-4\right)=1-4x^2-x^3+4x\)
\(=\left(1-x^4\right)+4x\left(1-x\right)=\left(1-x\right)\left(x^2+x+1\right)+4x\left(1-x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
x(x+2)(x^2+2x+2)+1
(x^2+2x)(x^2+2x+2)+1
dat x^2+2x=a
=> a(a+2)+1
=a^2+2a+1
=(a+1)^2
=(x^2+2x+1)^2
=(x+1)^4
x(x+2)(x^2+2x+2)+1
(x^2+2x)(x^2+2x+2)+1
dat x^2+2x=a
=> a(a+2)+1
=a^2+2a+1
=(a+1)^2
=(x^2+2x+1)^2
=(x+1)^4
1)x^2+x+2013x+2013=x(x+1)+2013(x+1)=(x+1)(x+2013)