I X--2 I--4=0 ( I 绝对值符号)

I X--2 I--4=0 ( I 绝对值符号) 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 2X--4 I=6 ( I 绝对值符号) x²-(4-i)x+5-5i=0 x²-(4-i)x+5-5i=0 lingo中这个INVALIDSETNAME.错误怎么改model:sets:bzx/1..7/:x,y,t,w,n;endsetsmin=20.4*10^2-@sum(bzx(i):(t(i)*x(i)+t(i)*y(i)));@for(bzx(i):@gin(x(i))); @for(bzx(i):@gin(y(i)));@for( @sum(bzx(i)|i#gt#4:t(i)*x(i)+t(i)*y(i)) 0-1规划用lingo编程问题@sum(place(i)|i#le#4:x(i)) > 2 @sum(place(i)|i#ge#4:x(i)) > 2 ;@sum(place(i)|i#eq#4:x(i)) > 2 下面的一元五次方程怎么解呢?最终i= 313.63X[ (1+i)^-1]+640.16X[ (1+i)^-2]+1015.24X[ (1+i)^-3]+1203.43X[ (1+i)^-4]+1576.53X[ (1+i)^-5]–275 = 0 为什么(2-3i)x(1+i)=5-i (1+2i)(x+yi)=4+3i (x-2y)+(y+2x)i=4+3i (i=2;i 在复数范围内解方程:x^2-(2+2i)x+4i=0 1.28 C语言,int x[3][2]={0},i;for(i=0;i 已知,I x I+I 3y-8 I+I z-2 I = 0,求代数式x+3y+z的值?已知,I x I+I 3y-8 I+I z-2 I = 0,求代数式x+3y+z的值? 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( X=[X;fliplr(x(i:2*N+i))] c++基础题一道#include void main() { int i=0,x=0,y=0; do{ ++i; if(i%2!=0) {x=x+i;i++;} y=y+i++; }while(i X^2+(4+i)+(3-3i)=0 的虚数根