I1-√2I+I√2-√3I+I√3-2I+I2-√5I+.+I√2010-√2011I

I√2-√3l-I1-√3I-I1+√2I I1-√2I+I√2-√3I+I√3-2I+I2-√5I+.+I√2010-√2011I 求出式子 I1/3-1/2I+I1/4-1/3I+I1/5-1/4I+.+I1/100-1/99I （I I 表示绝对值） I1/3-1/2I+I1/4-1/3I+I1/5-1/4I+…+I1/2010-1/2009I= 计算或化简：（1）I1－√2I＋I√2－√3I＋I√3－2I；（2）³√6﹣π﹣2（精确到0.01）I1－√2I＋I√2－√3I＋I√3－2I；（2）³√6﹣π﹣2（精确到0.01） 初中题l1/3-1/2I+I1/4-1/3I+I1/5-1/4I+.+I1/2005-1/2004I I1/3—1/2I+I1/4—1/3I+I1/5—1/4I+...+I1/10—1/9I的解你回吗?I I就是绝对直的意思 求式子I1/3-1/2I+I1/4-1/3I+I1/5-1/4I+.+I1/10-1/9I=I1/5-150/559I+I150/559-1/3I-I-1/3I用合理的方法算 I1/3-1/2I+I1/4-1/3I+I1/5-1/4I+……+I1/2009-1/2008I 火速.先回答加分.先30.列式+道理再+302小时内回答 (-1+√3i)^3/(1+i)^6+(-2+i)/(1+2i)的值是 ((-1+√3i)^3/(1+i)^6)-((-2+i)/(1+2i))过程怎么写 计算((-1+√3i)^3/(1+i)^6)+((-2+i)/(1+2i)) 已知i1=5√2sin(wt+30)A,和i2=5√2sin(wt-30)A,求（1）i=i1+i2,(2)i=i1-i2, [(1-√3i)^5-(1+√3i)^4]/[i*(-1+i^8)*(1/2+1/2i)]的答案是2√3-4i, (3i/√2-i)^2的虚部是多少 （x-1-√2i)(x-1+√2i)(x-2+√3i)(x-2-√3i) z=(3+4i)(√2-√2i)/(1/2-√3i)(-√3+i)求模长 复数的运算 计算（2+5i)÷3i+(√3+i)(√3-i)求解答步骤