求解七元一次方程组,34.04x1+54.96x2+14.22x3+9.03x4+53.46x5+x6-1.5507x7

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/09 12:25:43

求解七元一次方程组,
34.04x1+54.96x2+14.22x3+9.03x4+53.46x5+x6-1.5507x7=0
34.66x1+54.83x2+13.99x3+9.03x4+55.00x5+x6-1.5613x7=0
34.39x1+56.29x2+14.09x3+9.03x4+54.33x5+x6-1.5678x7=0
33.89x1+54.40x2+14.48x3+8.98x4+54.62x5+x6-1.5500x7=0
34.41x1+57.20x2+14.19x3+8.87x4+56.51x5+x6-1.5808x7=0
34.68x1+59.00x2+14.48x3+9.07x4+58.02x5+x6-1.6111x7=0
34.91x1+56.86x2+14.40x3+9.08x4+57.25x5+x6-1.5935x7=0
以上x1,x2,---x7,都是未知数,请别误认作乘号.请帮忙算算,

-2M1
2X-Y=5+M 所以4X-2Y=10+2M 所以4X-2Y++X+2Y=10+2M+3M 5X=10+5M X=2+M 2+M0 M-2
2+M+2Y=3M Y=M-1 M-10 M1 所以-2M1

求解七元一次方程组,34.04x1+54.96x2+14.22x3+9.03x4+53.46x5+x6-1.5507x7 有整数x1,x2,x3,x4,x5,x6,x7.x1 求解最佳方案X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 设x1~x7是自然数,且x1＜x2＜...＜x7,x1+x2=x3,x2+x3=x4,x3+x4=x5,x4+x5=x6 已知x1+x2+x3+x4+x5+x6+x7=2010,且x1+x2=x3,x2+x3=x4,x3+x4=x5,x5+x 设x1,x2,x3,x4,x5,x6,x7是自然数,且x1 x1,x2,x3,x4,x5,x6,x7为自然数,并且x1 如何计算x1,x2,x3,x4,x5,x6,x7的最小整数值 Min Z= X1 +X2 +X3 +X4 +X5 +X6 +X7 +X8 一道数学题,很难啊x1,x2,x3,x4,x5,x6,x7是七个有理数,且x1小于x2小于x3小于x4小于x5小于x6小 x8+x7+x6+x5+x4+x3+x2+x+1因式分解设x1，x2，x3，x4，x5，x6，x7是自然数，且x1＜x2＜x3＜x4＜x5＜x6＜x7，x1+x2=x3，x2+