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答案:
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1.1.设复数z满足(1 i)z=2i,则z=()
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2.1.已知/k的z变(z )(z 2z)的收敛域为[k是因果序列A、|z|2C、<|z|<2
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3.X(z)=s(s 1)X(z)=S 13X(z)=1-e-ss2(s 1
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4.1.设z=fnx ,其中函数f(u)可微则 yax
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5.设 z = f(u, v),u = φ(x, y),v = ψ(x, y),则∂z/∂x = ( )
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6.A.Z <ZαB.Z >ZαC.Z <-ZαD.Z >-Zα
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7.hi=1 z
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8.{u,v,w}/.u→v/.v→z? {w,z,w}{z,z,w}{w,w,z}{z,w,w}
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9.14 lion(1 z)-
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10.函数yA、[ ] (k Z)B、[ ] (k Z) C . [ ] (k Z) D. [ ] (k Z)