Question

数学高手来解答，求答案，下面的数学题，急求正确率，给高分，做好之后拍照贴上来，谢谢！！！第二次作业

Answer 1

10.
(1)原式=√2e^(πi/4)×2e^(πi/3)÷e^(πi/6)=√2×2×e^(πi/4+πi/3-πi/6)=2√2e^(5πi/12)=2√2[cos(5π/12+isin(5π/12)]=(√3-1)+(√3+1)i
(2)原式=[2e^(πi/4)]^5×[√3e^(πi/6)]^6÷[√6e^(3πi/2)]²=[2^5×(√3)^6÷(√6)²]×e^(5πi/4+πi-3πi)=144e^(-3πi/4)=144[cos(3π/4)-isin(3π/4)]=-72√2+72√2i
(3)原式=[cos(π/6)+isin(π/6)]^6+[cos(π/6)-isin(π/6)]^6=[e^(πi/6)]^6+[e^(-πi/6)]^6=e^(πi)+e^(-πi)=(cosπ+isinπ)+(cosπ-isinπ)=-2
(4)原式=[2e^(5πi/3)]^4÷[e^(πi/3)]^3=16×e^[20πi/3-πi]=16e^(17πi/3)=16[cos(17π/3)+isin(17π/3)]=16[cos(-π/3)+isin(-π/3)]=8-8√3i

11.
x²+7=(x+√7i)(x-√7i) 2x²-6x+5=(x-3/2+i/2)(x-3/2-i/2)

12.
(1)证明：因为ω=-1/2+√3i/2=cos(2π/3)+isin(2π/3)=e^(2πi/3)，所以ω³=[e^(2πi/3)]³=e^(2πi)=cos(2π)+isin(2π)=1，即ω³-1=(ω-1)(ω²+ω+1)=0。
因为ω≠1，所以ω²+ω+1=0
(2)证明：由(1)得：1+ω=-ω²、1+ω²=-ω，故(1+ω-ω²)³-(1-ω+ω²)³=(-2ω²)³-(-2ω)³
因为ω³=1，所以(-2ω²)³-(-2ω)³=(-2)³-(-2)³=0，即(1+ω-ω²)³-(1-ω+ω²)³=0

Answer 2

10、（1）√2﹙cosπ/4＋isinπ/4﹚×2﹙cosπ/3＋isinπ/3﹚÷﹙cosπ/6＋isinπ/6﹚
＝√2﹙√2/2＋√2i/2﹚×2﹙1/2＋√3i/2﹚÷﹙√3/2﹢i/2﹚
=2﹙1＋i﹚﹙1＋√3i﹚/﹙√3＋i﹚
＝2﹙1＋√3i＋i＋√3i²﹚/﹙√3＋i﹚
=2[1－√3＋﹙√3＋1﹚i]/﹙√3＋i﹚
=√3－1＋﹙√3＋1﹚i
题目太长我就不抄一遍了,下面五次方六次方你将就看吧
﹙2﹚=[2﹙√2/2＋√2i/2﹚﹚]^5×[√3﹙√3/2＋i/2﹚]^6÷[√6﹙0＋i﹚]^2
=﹙√2＋√2i﹚^5×﹙3/2＋√3i/2﹚^6÷6i^2
=4√2﹙1＋i﹚^5×﹙3＋√3i﹚^6/64÷﹙﹣6﹚
=﹣√2﹙1＋i﹚^5×﹙3＋√3i﹚^6/96
=﹣√2﹙1＋2i＋i^2﹚^3×﹙9＋6√3i＋3i^2﹚^4/96
=﹣√2×8i^3×﹙6＋6√3i﹚^4/96
=√2i﹙36＋72i+108i^2﹚^2/12
=√2i﹙﹣72+72i﹚^2/12
=√2i﹙5184-10368i+5184i^2﹚/12
=-10368√2i^2/12
=864√2
(3)=[﹙√3＋i﹚^6＋﹙√3－i﹚^6]/64
=[﹙3＋2√3i＋i^2﹚^4＋﹙3－2√3i＋i^2﹚^4]/64
=[﹙4＋8√3i＋4i^2﹚^2＋﹙4-8√3i＋4i^2﹚^2]/64
=﹙192i^2＋192i^2﹚/64
=-6
11、
（1）x²＋7
＝x²－7i²
＝﹙x＋√7i﹚﹙x－√7i﹚
(2)2x²－6x＋5
＝﹙x²-4x＋4﹚＋﹙x²-2x＋1﹚
＝﹙x-2﹚²＋﹙x-1﹚²
＝﹙x-2﹚²－﹙x－1﹚²i²
＝[x－2＋﹙x-1﹚i][﹙x－2－﹙x－1﹚i]
12、ω=－1/2＋√3i/2
（1）1＋ω＋ω²
＝1＋﹙－1/2＋√3i/2﹚＋﹙－1/2＋√3i/2﹚²
＝1－1/2＋√3i/2＋1/4－√3i/2＋3i²/4
＝1－1/2＋√3i/2＋1/4－√3i/2－3/4
＝0
﹙2﹚﹙1＋ω－ω²﹚³－﹙1－ω＋ω²﹚³
=﹙1＋ω＋ω²－2ω²﹚³－﹙1＋ω＋ω²－2ω﹚³
＝﹙－2ω²﹚³－﹙－2ω﹚³
＝－8[﹙－1/2＋√3i/2 ﹚²]³＋8﹙－1/2＋√3i/2﹚³
＝－8﹙1/4－√3i/2＋3i²/4﹚³＋8﹙－1/2＋√3i/2﹚³
＝8﹙1/2＋√3i/2﹚³＋8﹙－1/2＋√3i/2﹚³
＝8[﹙1/2＋√3i/2﹚³＋﹙－1/2＋√3i/2﹚³]
＝8﹙1/2＋√3i/2－1/2＋√3i/2﹚[﹙1/2＋√3i/2﹚²－﹙1/2＋√3i/2﹚﹙－1/2＋√3i/2﹚＋﹙－1/2＋√3i/2﹚²
＝8√3i[﹙1/4＋√3i/2＋3i²/4﹚－﹙－1/4＋3i²/4﹚＋﹙1/4－√3i/2＋3i²/4﹚]
＝8√3i×0
＝0
补充说明：^后面加数字表示次方，^4四次方，^5五次方，^6六次方。
i²＝﹣1
立方和公式：
a³﹢b³＝﹙a﹢b﹚﹙a²-ab﹢b²﹚

Answer 3

我上课那会儿没专心，这些题目我都不会做啊

Answer 4

脖子疼