多项式已知Px=z^n c1z^n-1是复变量z的实系数
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C=-A-B=-(x²+2y²-z)-(-4x²+3y²+2z)=-x²-2y²+z+4x²-3y²-2z=3x
当x=-2时,y=m•(-2)5+n•(-2)3+p(-2)-7=5,则-25m-23n-2p-7=5,-25m-23n-2p=12,当x=2时,y=25m+23•n+2p-7,两式相加:y=-12-
答:3x^2-px-2=(x+q)(3x-1)3x^2-px-2=3x^2-x+3qx-q-px-2=(3q-1)x-q所以:-p=3q-1q=2解得:p=-5,q=2再问:我的问题是3x²
不妨用特殊代入法啊令a=b=c=0或者a=1,b=-1,c=0结果都是x^3+x^2z-xyz+y^3=0
设px2−yz=qy2−zx=rz2−xy=k,则p=(x2-yz)k,q=(y2-zx)k,r=(z2-xy)k.已知p+q+r=9,则(x2-yz)k+(y2-zx)k+(z2-xy)k=9,即k
题目问题应该是求x=-2时的值吧当x=2时,mx^5+nx^3+px-4=32m+8n+2p-4=532m+8n+2p=9当x=-2时,mx^5+nx^3+px-4=-32m-8n-2p-4=-(32
∵一元二次方程x2+px+q=0的两根为a,b,∴由根与系数关系知,a+b=-p,ab=q,∴原方程化为x2-(a+b)x+ab=0,∴多项式x2+px+q可因式分解为(x-a)(x-b),故选A.
(I)当p=2时,函数f(x)=2x-2x-2lnx,f(1)=2-2-2ln1=0.f′(x)=2+2x2-2x,曲线f(x)在点(1,f(1))处的切线的斜率为f'(1)=2+2-2=2.从而曲线
设a=2009A,b=2008则x=a+b-1,y=a+b,z=a+b+1x^2+y^2+z^2-xy-yz-xz=(a+b-1)^2+(a+b)^2+(a+b+1)^2-(a+b-1)(a+b)-(
c=-A-B=-x^2-2y^2+z^2+4x^2-3y^2-2Z^2=3x^2-5y^2-z^2
题目中的A里,y应该也有平方的吧,我就按有平方的做了C=-A-B=-(x²+2y²+z²)-(-4x²+3y²+2z²)=-x²-
A+B+C=0C=-A-B=-(x²+2y²-z²)-(-4x²+3y²+2z²)=-x²-2y²+z²+4x
y=3xz=4y=12xx-2y+3z=62x-6x+36x=6231x=62x=2y=6z=24xy-yz+xz=y(x-z)+xz=48-6*22=48-132=-84
P=-3M-2N=-3x-9y+9z+4x-10y+2z=x-19y+11z
(x-y)²+(x+z)²-(y+z)²=4+(x+z)²-9=+(x+z)²-5此时,把题目再处理一下得出:X+z=5则最后结果为25-5=20
x=2时,mx^5+nx^3+px-4=32m+8n+2p-4=532m+8n+2p=9-(32m+8n+2p)=-9x=-2时mx^2+nx^3+px-4=-32m-8n-2p-4=-(32m+8n
-13首先把2带入原式得m2^5+n2^3+2p-4=5【1】然后把-2带入原式得-m2^5-n2^3-2p-4【2】由【1】可得-m2^5-n2^3-2p=-9【3】把【3】带入【2】即得-13
mx^5+nx^5+px-4=(m+n)x^5+px-4x=2,2^5(m+n)+2p-4=532(m+n)+2p=9,两边同乘-1得-32(m+n)-2p=-9x=-2,原式=-32(m+n)-2p