在复数范围内分解因式3x^2-6x 4
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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)
平方差公式!x^4+y^4=(x²+iy²)(x²-iy²)=(x+i√iy)(x-i√iy)(x+√iy)(x-√iy)
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
x²+2x+3=x²+2x+1+2=(x+1)²+2=(x+1+√2i)(x+1-√2i)再问:倒数第二步是怎么变成最后一步的?再答:平方差
2x^2-4x+3=(√2x)^2-2(√2x)(√2)+(√2)^2+1=(√2x-√2)^2-[√(-1)]^2=(√2x-√2)^2-i^2=(√2x-√2+i)(√2x-√2-i)
x^3+2x^2+3x+6=x^2(x+2)+3(x+2)=(x+2)(x^2+3)=(x+2)(x+根号3i)(x-根号3i)
x^2-2x+3=(x-1-根号2i)(x-1+根号2i)
1.x^4-4y^4=(x^2+2y^2)(x^2-2y^2)=(x+根号2yi)(x-根号2yi)(x+根号2y)(x-根号2y)2.-1/2x^2+x-3=-1/2(x^2-2x+6)=-1/2[
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))
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-1)(x-2);(3y-4)的平方;(x平方-3)(x平方+1)
x^3-2x=x(x^2-2)(有理数范围内)=x(x+√2)(x-√2)(实数范围内)
原式=(x^5-1)/(x-1)先求出x^5-1=0的根,再除去1这个根即可表示由x^5-1=0知,x为5次单位圆根,故x1=1,x2=cosa+sinai,x3=cos2a+sin2ai,x4=co
2x^2+2x+3设2x^2+2x+3=0,解之得x=[-1±(√5)i]/2,所以原式={2x-[-1+(√5)i]}{x-[-1-(√5)i/2}=[2x+1-(√5)i]{x+1/2+(√5)i
x^3-1=(x-1)(x²+x+1)=(x-1)[x+(1+√3i)/2][x+(1-√3i)/2]
原式=3(x²+x/3-1/3)=3(x²+x/3+1/36-1/36-1/3)=3[(x+1/6)²-13/36]=3(x+1/6+√13/6)(x+1/6-√13/6
3x^2-x=x(3x-1)(提取公因式x)