(1+0.4567+0.5678)×(0.4567+0.5678+0.6789)－(1+0.4567+0.5678+0.6789)×(0.4567+0.5678）＝
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(1+0.4567+0.5678)×(0.4567+0.5678+0.6789)－(1+0.4567+0.5678+0.6789)×(0.4567+0.5678）＝(1+0.4567+0.5678)×
(1+0.4567+0.5678)×(0.4567+0.5678+0.6789)－(1+0.4567+0.5678+0.6789)×(0.4567+0.5678）＝
(1+0.4567+0.5678)×(0.4567+0.5678+0.6789)－(1+0.4567+0.5678+0.6789)×(0.4567+0.5678）＝
(1+0.4567+0.5678)×(0.4567+0.5678+0.6789)－(1+0.4567+0.5678+0.6789)×(0.4567+0.5678）＝
设0.4567+0.5678=t
则：
原式=(1+t)(t+0.6789)-(1+t+0.6789)*t
=t(1+t)+0.6789(1+t)-t(1+t)-0.6789t
=0.6789
0.6789
设（0.4567+0.5678）为x
则原式=（1+x)x(x+ 0.6789)-（1+x +0.6789) ×(x）
=x+ 0.6789+x^2+ 0.6789x-x-x^2－0.6789x
= 0.6789
=(1+0.4567+0.5678)*(0.4567+0.5678)+(1+0.4567+0.5678)*0.6789-(1+0.4567+0.5678)*(0.4567+0.5678)-0.6789*(0.4567+0.5678)=(1+0.4567+0.5678)*0.6789-0.6789*(0.4567+0.5678)=0.6789+0.6789*(0.4567+0.5678)-0.6789*(0.4567+0.5678)=0.6789
(1+0.4567+0.5678)×(0.4567+0.5678+0.6789)－(1+0.4567+0.5678+0.6789)×(0.4567+0.5678）＝
0.5678 678循环化分数是?
（0.2345*3456-1234*0.4567）/【（6666²-5555²）*4444】
（0.2345*3456-1234*0.4567）/【（6666²-5555²）*4444】
如何用matlab求下面邻接矩阵的可达矩阵A=[0 0 0 0 0 0 0;1 0 0 0 0 0 0;1 1 0 1 0 0 0;1 0 0 0 0 0 0;1 1 0 0 0 0 0;1 0 0 1 0 0 1;1 0 0 0 0 0 0];
矩阵求逆 1 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0
线性代数关于矩阵的问题?1 0 0 0 0 0 1 0 0 0 0 0 1 2 1 2 1 2 0 1 0 0 0 0 0 1 0 0 0 0 1 2 1 2 1 2 1 2 1 2 1 2 0 0 1 0 0 0 0 0 1 0 0 01 2 1 2 1 2 0 0 0 1 0 0 0 0 0 1 0 00 0 0 0 1 0 1 2 1 2 1 2 0 0 0 0 1 0 0 0 0 0 0 1 1 2 1 2 1 2 0 0 0 0 0 1 1
0+0=1
8 0 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 7 0 5 0 0 6 0 0 1 0 0 0 4 9 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 7 0 9 0 0 5 0 0 2 4 0 8 0 0 0 0 0
求通项公式：1,0,-1,0,1,0,-1,0,...
1,0
matlab解线性方程组 --------------------------------------------------------------------------------A=[0 0 1 0 0 1 0 1 0 0 700;0 1 -1 1 0 0 0 0 0 0 300; 0 0 0 1 1 0 0 0 0 0 500;0 0 0 0 0 -1 1 0 0 0 200;1 1 0 0 0 0 0 0 0 0 800; 1 0 0 0 1 0 0 0 0
matlab SVD分解结果X=[1 0 0 1 0 0 0 0 01 0 1 0 0 0 0 0 01 1 0 0 0 0 0 0 00 1 1 0 1 0 0 0 00 1 1 2 0 0 0 0 00 1 0 0 1 0 0 0 00 0 1 1 0 0 0 0 00 1 0 0 0 0 0 0 10 0 0 0 0 1 1 1 00 0 0 0 0 0 1 1 10 0 0 0 0 0 0 1 1];[T,S,D]=svd(X,0)分解后得到的
1,0,1,0,1,0,1,0,1.0,1,0,1,0,1,0,1,0.
0 1 1 0 0 0 0 0 0 这个矩阵的基础解析怎么看0 1 1、0 0 0、0 0 0
独立事件会发生的概率0 0 1 0 2 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 2 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 1 0 0 0 1 2 0 0 0 0 0 0 1 1 0 0 0 2 0 1 0 0 0 0 0 0 2 2 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0
matlab创建m文件Num0=[1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 1 1 1 1];Num1=[0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0];Num2=[1 1 1 1 1
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