设z=e-x一f(x一2y),且当y=0时,z=x2,求

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问题 设z=e-x一f(x一2y),且当y=0时,z=x2,求

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答案由于y=0时,z=x2,而z=e—x一f(x一2y).当y=0时,可得z=e—x—f(x),则 e—x—f(x)=x2, f(x)=e—x一x2, 即 f(x一2y)=e—(x—2)一(x一2y)2, z=e—x一e2y—x+(x一2y)2. 故 [*]=e2y—x一e—x+2(x一2y).

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