(1+i)x ^2-(3-i)x+(2-2i)=0,1）求该方程有没有实数根2）求出该方程所有的根△求出来是个虚数啊,怎么办.虚数又不能比较大小的

为什么(2-3i)x(1+i)=5-i （x-1-√2i)(x-1+√2i)(x-2+√3i)(x-2-√3i) (1+2i)(x+yi)=4+3i (x-2y)+(y+2x)i=4+3i 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 已知(x/1-i)+(y/1-2i)=5/1-3i求x+y For i=−1,if 3i(2+5i)=x+6i,then x=? 虚数 化简i^(7x+ 2)还有i^(6x-1),i^(14x-11) |X/i^2| 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 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 已知2＜x＜3,那么I x-根号2 I+I x-根号10 I=? （1+2i）x+(3-10i)y=5-6i X=[X;fliplr(x(i:2*N+i))] 已知x,y为实数,i为虚数单位.若x/(1-i)+y/(1-2i)=5/(1-3i),则x+y的值为? 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平方+(1-3i)x+(2-3i)=0 x,y∈R,x/(1-i) -y/(1-2i)=5/(1-3i)则xy=( ) Var x=3,y=2; Int I,sum=5; If (x>y) {for (i=1;i