x4次方加1因式分解 x的四次方加1因式分解

x^4+1=x^4+2x²+1-2x²=(x²+1)²-(√2x)²=(x²+√2x+1)(x²-√2x+1) 把一个多项式在一个范围(如实数范围内分解,即所有项均为实数)化为几个整式的积的形式,这.

16x^4-1=(4x²+1)(4x²-1)=(4x²+1)(2x+1)(2x-1)

4x的4次方加1=4x^4+4x^2+1-4x^2=(2x^2+1)^2-(2x)^2=(2x^2+2x+1)(2x^2-2x+1)

解:4·x的4次方 + 1 =(4·x的4次方 + 4·x平方 + 1) - 4·x平方 =(2·x平方 + 1)平方 - (2x)平方 =(2·x平方 +2x+1)(2·x平方 -2x +1)【很高兴为你解决以上问题,希望对你的学习有所帮助!】

X4+1=x4+2x²+1-2x²=(x²+1)-2x²=(x²+√2*x+1)(x²-√2*x+1)

x的四次方-1=x²(x²-1)=x²(x+1)(x-1)

x^4+1 解:添项,得:x^4+2x^2+1-2x^2=(x^2+1)^2-(√2*x)^2=(x^2+1+√2*x)(x^2+1-√2*x)=(x^2+√2*x+1)(x^2-√2*x+1)

4x^4+1=4x^4+4x^2+1-4x^2=(2x^2+1)^2-(2x)^2=(2x^2-2x+1)(2x^2+2x+1)

x的4次方+1 =x的4次方+1-2x*x+2x*x =(x*x+1)(x*x+1)-2x*x =(x*x+√2x+1)(x*x-√2x+1)

x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(√2x)^2=(x^2+1+x√2)(x^2+1-x√2)