在复数集中分解因式matlab
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△=16-40=-24x1,2=2±i根号6/2故2x²-4x+5=(x-(2+i根号6/2))(x-(2-i根号6/2))
x^3-2x^2+4x-8=x^2(x-2)+4(x-2)=(x-2)(x^2+4)=(x-2)(x-2i)(x+2i)
x^3+1=x^3-(i)^3=(x-i)(x^2+xi-1)
x^2-2x3=(x-1-根号2i)(x-1根号2i)
令2x^2-4x+5=0x1=[4+√(16-40)]/4=1+√6i/2x2=1-√6i/22x^2-4x+5=2(x-1-√6i/2)(x-1+√6i/2)
x^6-y^6=(x^3+y^3)(x^3-y^3)=(x+y)(x^2-xy+y^2)(x-y)(x^2+xy+y^2)=(x+y)(x-y)[x-(根号3)yi][(x+(根号3)yi][3x-(
x∧2-y∧2=(x+y)(x-y)再问:x-y在复数范围内还能分吧再问:x-y在复数范围内还能分吧再答:=(x+y)(x-y)==(x+y)(√x+√y)(√x-√y)没有常数,没办法弄出复数i
平方差公式!x^4+y^4=(x²+iy²)(x²-iy²)=(x+i√iy)(x-i√iy)(x+√iy)(x-√iy)
2X^2+X+1=2[x-(-1+√7i)/4][x-(-1-√7i)/4]
a^3+8b^3=a^3+(2b)^3=(a+2b)[a^2-2ab+(2b)^2]=(a+2b)[(a^2-2ab+b^2)+b^2]=(a+2b)[(a-b)^2-(bi)^2]=(a+2b)(a
2x"-4x+5=2x"-4x+2+3=2[(x-1)"+3/2]=2[(x-1)"-(-6/4)]=2{(x-1)"-[(√6/2)i]"}=2[x-1-(√6/2)i][x-1+(√6/2)i]或
楼上的因式分解可以把√(2i)ix⁴+4=(x²+2i)*(x²-2i)假设(a+bi)²=2i,得a²-b²+2abi=2i,则a
原式=(a²+1)(a²-1)=(a+1)(a-1)(a²+1)如果本题有什么不明白可以追问,如果满意请点击“选为满意答案”再问:有一个空心的银制小球,外直径为12cm,
2X^2+4X+2+2=2(X^2+2X+1-1)=2[(X+1)^2+1]
x^2-2x+3=(x-1-根号2i)(x-1+根号2i)
2x²+3x+2=2[x-﹙-3+√7i﹚/4][x-﹙-3-√7i﹚/4]=2[x+﹙3-√7i﹚/4][x+﹙3+√7i﹚/4]
=(x-1)(x^2+x+1)=(x-1)(x+1/2*(1+根号3i))(x+1/2*(1-根号3i))
x^4-4=(x^2+2)(x^2-2)=(x+√2i)(x-√2i)(x+√2)(x-√2)如果本题有什么不明白可以追问,请点击下面的【选为满意回答】按钮,
首先,在复数范围内解方程x^2+4=0,求的x1=2i,x2=-2i,则x^2+4=(x+2i)(x-2i)
x^3-1=(x-1)(x²+x+1)=(x-1)[x+(1+√3i)/2][x+(1-√3i)/2]