解方程组x1 x2=x2 x3
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n(n+1)(n+2)(n+3)/4
解:令x1=y1+y2,x2=y1-y2,x3=y3,x4=y4f=y1^2-y2^2+y1y3-y2y3+y3y4=(y1+y3/2)^2-(y2+y3/2)^2+y3^2y3y4=z1^2-z2^
由x1x2x3…x2007=x1-x2x3…x2007=x1x2-x3…x2007=…=x1x2x3...x2006-2007=1可知:x1x2x3...x2006-1/x1x2x3...x2006=
因为x1,x2,x3是原方程的三个根,所以,原方程可写作:(x-x1)(x-x2)(x-x3)=0解开得:x^3-(x1+x2+x3)x^2+(x1x2+x2x3+x1x3)x-x1x2x3=0而原等
A=011101110A+E=111111111-->111000000对应方程x1+x2+x3=0(1,-1,0)^T显然是一个解与它正交的解有形式(1,1,x)^T代入方程x1+x2+x3=0确定
A=1-22-2-24240嗯,特征值好麻烦-6074/97723143/977估计题目有误.
nx(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)]1x2+2x3+3x4+...+nx(n+1)=1/3[1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+
读完以上材料,请你计算下列各题:(1)1×2+2×3+3×4+…+10×11(写出过程);1×2+2×3+3×4+…+10×11=1/3(1×2×3-0×1×2)+1/3(2×3×4-1×2×3)+1
3*(1x2+2x3+3x4+...+99x100)=3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100)=99x100x1
1X2+2X3+3X4+、、、、、、+nX(n+1)=(1/3)(1*2*3-0*1*2)+(1/3)(2*3*4-1*2*3)+(1/3)(3*4*5-2*3*4)+.+(1/3)[n*(n+1)(
解:二次型的矩阵A=1-24-242421|A-λE|=1-λ-24-24-λ2421-λ=-(λ+4)(λ-5)^2A的特征值为λ1=-4,λ2=λ3=5.对λ1=-4,(A+4E)X=0的基础解系
f=(x1-2x2+2x3)^2-6x2^2-6x3^2+16x2x3=(x1-2x2+2x3)^2-6(x2-4/3x3)^2+(14/3)x3^2令(y1,y2,y3)'=(x1-2x2+2x3,
(1)A=11010-10-11(2)|A-λE|=1-λ101-λ-10-11-λc1+c31-λ100-λ-11-λ-11-λr3-r11-λ100-λ-10-21-λ=(1-λ)[-λ(1-λ)
∵方程(x-1)(x2+8x-3)=0的三根分别为x1,x2,x3,∴x1=1,x3+x2=-8,x3•x2=-3,则x1x2+x2x3+x3x1=x1(x2+x3)+x2x3=-3-8=-11.故选
令x1=y1+y2,x2=y1-y2,x3=y3则f=2(y1+y2)(y1-y2)+2(y1+y2)y3-6(y1-y2)y3=2y1^2-4y3y1-2y2^2+8y3y2=2(y1-y3)^2-
(1)1x2+2x3+…+99x100+100x101==1/3x100x101x102=343400(2)1x2+2x3+3x4+…+n(n+1)(n为正整数)=1/3n(n+1)(n+2)(3)1
(1)1x2+2x3+3x4+…+10x11=1*10*11*12/3=440(2)原式=n(n+1)(n+2)/3(3)1x2x3+2x3x4+3x4x5+…+7x8x9=1x2x3×4/4-0×1
3*(1x2+2x3+3x4+...+99x100)=3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100)=99x100x1
∵f(x)是奇函数∴f(-x)=-f(x)∴对任意的x,有-x³+(b-1)*(-x)²+c*(-x)=-x³-(b-1)x²-cx化简,得2(b-1)x
f(x1,x2,x3)=2x1x2+2x1x3+2x2x3对应的实对称矩阵为A=[(0,1,1)T,(1,0,1)T,(1,1,0)T];下面将其对角化:先求A的特征值,由|kE-A|=|(k,-1,