几道微积分必学基础题请写出 标准式 （standard form 不知道中文是不是这个）6/i 2/4-5i(2+i)/(2-i)i/(4-5i)^2perform the operation and write the result in standard form2/(1+i)-3/(1-i)请写出步骤! 因为我需要步骤去自习!我给100分! 十万火急! 谢谢!

6/i=-6*i^2/i=-6i
2/4-5i=2(4+5i)/(4-5i)(4+5i)=(8+10i)/41
(2+i)/(2-i)=(2+i)(2+i)/(2+i)(2-i)=(3+4i)/5
i/(4-5i)^2 =i(4+5i)^2/[(4+5i)^2*(4-5i)^2]
主要思想是分子实数化，相信你一定能理解^-^

6/i=-6*i^2/i=-6i
2/4-5i=2(4+5i)/(4-5i)(4+5i)=(8+10i)/41
(2+i)/(2-i)=(2+i)(2+i)/(2+i)(2-i)=(3+4i)/5
i/(4-5i)^2 =i(4+5i)^2/[(4+5i)^2*(4-5i)^2]

这不是微积分吧？这是复数。
标准式就是把复数写成a+bi的形式。
自己算吧。注意i的平方=-1就行了。

6/i =-6I
2/4-5i =1/2-5I
(2+i)/(2-i) =i/(4-5i)^2
perform the operation and write the result in standard form
2/(1+i)-3/(1-i)

6/i
=6i/(i)^2
=-6i
(i)^2=-1
2/(4-5i)
=2(4+5i)/[(4-5i)(4+5i)]
=(8-10i)/(16+25)
=(8-10i)/41
(2+i)/(2-i)
=(2+i)^2/[(2+i)(2-i)]
=(4+2i-1)/5
=(3+2i)/5
i/(4-5i)^2
=i/(16-40i-25)
=i/(-40i-9)
=-i/(9+40i)
=-i(9-40i)/[(9+40i)(9-40i)]
=(-40-9i)/(81+40)
=-(40+9i)/121
2/(1+i)-3/(1-i)
=2(1-i)/2-3(1+i)/2
=[2-2i-3-3i]/2
=(-1-5i)/2
一般:
k/(a+bi)
=k(a-bi)/[(a+bi)(a-bi)]
=k(a-bi)/(a^2+b^2)