I x-3 I+I x-m I < 5 求m的取值范围求结果+过程,我记得有个公式可以直接代的

为什么(2-3i)x(1+i)=5-i |X+i| |x-i| I 1 3 0 I I 7 I I 0 1 -1 I X= I 2 I求X.大哥大姐,用矩阵的初等变换写~I 2 1 5 I I 4 I I x-3 I+I x-m I < 5 求m的取值范围求结果+过程,我记得有个公式可以直接代的 matlab Index exceeds matrix dimensions.子函数：function [MminT] =MminT(M,H,r,i)syms x;y=(x-i(M)).*(x-i(M+1))./((i(M-1)-i(M)).*(i(M-1)-i(M+1))).*r(M-1,H)+(x-i(M-1)).*(x-i(M+1))./((i(M)-i(M-1)).*(i(M)-i(M+1))).*r(M,H)+(x-i(M-1)).*(x-i(M))./((i(M+1) I-xI=I-3I,则x= 已知(x/1-i)+(y/1-2i)=5/1-3i求x+y fortran程序,subroutine solvereal::i,x,y,mreal::h(i),d(i),a(i),dy(i)h(i)=(1+2*(x-a(i))/(a(i+1)-a(i)))*(((x-a(i+1))/(a(i)-a(i+1)))**2)d(i)=(((x-a(i+1))/(a(i)-a(i+1)))**2)*(x-a(i))m=y(i)*h(i)+dy(i)*d(i)end subroutine solveprogram mainreal::i,sreal::a( For i=−1,if 3i(2+5i)=x+6i,then x=? 下面这个程序有什么错?为什么运行出来的z(i,j)总是为0呢? lingosets:m/1..6/:;n/1..5/:;ajs(m,n):x,y,z；endsetsmax=@sum(m(i):@sum(n(j)：y(i,j)*z(i,j)+x(i,j);@for(ajs:@bin(z));@for(m(i):z(i,1)+z(i,2)+z(i,3)+z(i,4)+z(i,5) I think I’m （1+2i）x+(3-10i)y=5-6i x²-(4-i)x+5-5i=0 x²-(4-i)x+5-5i=0 IaI有非负性,也有最小值0.试讨论:若x为有理数,I x I+I X-2 I+I X-3 I有没有最小值 （x-1-√2i)(x-1+√2i)(x-2+√3i)(x-2-√3i) matlab数组生成问题已知x为长度为M的一组数,X(i)=[x(i),x(i-1),...,x(i-M+1)]'怎样生成