根据2+4=6=2x3 2+4+6=12=3x4 2+4+6+8=20=4x5 2+4+6+8+10=30=5x6 计算 2a+4a+6a+.+100a的和根据2+4=6=2x3 2+4+6=12=3x4 2+4+6+8=20=4x5 2+4+6+8+10=30=5x6计算126a+128a+130a+132a.+300a的和
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/08 09:12:16
x){`uF&fF
@q makd`kRa
jؚV)X
F&fz@9ZNdF@3
,0J6`TOCPn<⼢_`g3b#OH
5un1525L/.H̳y >*B
解方程组x1+x2=4,x2+x3=-6,x3+x1=2
(5^4x3^3-5^3x3^2+5^2x3)÷15=?
写出下面线性规划的对偶规划min Z=3X1+2X2+X3; X1+X2+X3≤6;X1-X3≥4;X2-X3≥3;X1,X2,X3≥0
用初等行变换来解下列线性方程组(1)2x1-x2+3x3=3 3x1+x2-5x3=0 4x1-x2+x3=3 x1+3x2-13x3=-6(2) x1-2x2+x3+x4=1 x1-2x2+x3-x4=-1 x1-2x2+x3-5x4=5(3) x1-x2+x3-x4=1 x1-x2-x3+x4=0 x1-x2-2x3+2x4=-1/2
3x3÷2x3=
{2X1-X2+3X3=33X1+X2-5X3=04X1-X2+X3=3X1+3X2-13X3=-6{X1-2X2+3X3-4X4=4X2-X3+X4=-3X1+3X2-3X4=1-7X2+3X3+X4=-3
用配方法化二次型刘老师您好:f(x1,x2,x3)=x1^2+3x2^2+8x3^2+4x1x2+6x1x3+10x2x3配方后得f(x1,x2,x3)=(x1+2x2+3x3)^2-(x2+x3)^2 做可逆线性变换y1=x1+2x2+3x3 x1=y1-2y2-y3y2=x2+x3 即 x2=y2-y3y3=x3 x3=y3「即」前面的式
解线性方程组x1+x2+3x3+4x4=5 2x1+4x2+4x3+6x4=8 -x1-2x2-x3-2x4=-3
二次型f(x1,x2,x3)=(x1)^2+3(x2) ^2-4(x3)^2+6(x1)(x2)+10(x2)(x3)的矩阵是
{2X1 -X2 -X3 +X4 = 2{X1 +X2 -2X3 +X4 = 2{4X1 -6X2 +2X3 -2X4 = 4{X1 +4X2 -3X3 +2X4 = 1
求方程组X1-3X2-2X3-X4=1 3X1-8X2-4X3-X4=0 -2X1+X2-4X3+2X4=1 -X1-2X2-6X3+X4=2的解
求线性方程组{X1-3x2-2x3-X4=1;3X1-8X2-4X3-X4=0;-2X1+X2-4X3+2X4=1;-X1-2X2-6X3+X4=2的一般解.
求线性方程组 2 x1+3 x2+x3=4;x1-2 x2+4 x3=-5;3 x1+8 x2-2 x3=13;4 x1-x2+9 x3=-6的全部解
用列主元Gauss消元法解线性方程组{-x2-x3+x4=0,x1-x2+x3-3x4=1,2x1-2x2-4x3+6x4=-1,x1-2x2-4x3+x4=-1
消元法解线性代数2X1+3X2+ X3=4X1-2X2+4X3=-53X1+8X2-2X3=134X1-X2+9X3=-6
求线性方程组x1-3x2-2x3-x4=1,3x1-8x2-4x3-x4,-2x1+x2-4x3+2x4=1,-x1-2x2-6x3+x4=
解方程组:x1+x2+x3+x4+x5=7 3x1+2x2+x3+x4-3x5=-2 x2+2x3+2x4+6x5=23 5x1+4x2-3x3+3x4-x5=10
运筹学线性规划化标准型目标函数:min z=x1-x2-x3+2x4约束条件:10x1+x2-x3-4x4=77x1+6x2-2x3-5x4>=104x1-8x2+6x3+x4