设f(x)=(ax+b)sinx+(cx+d)cosx 试确定abcd的值,使f'(x)=xsinx-cosx 答案是c=1 b=-2 a=d=0而我是 c=-1 b=2 a=d=0
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![设f(x)=(ax+b)sinx+(cx+d)cosx 试确定abcd的值,使f'(x)=xsinx-cosx 答案是c=1 b=-2 a=d=0而我是 c=-1 b=2 a=d=0](/uploads/image/z/7623645-69-5.jpg?t=%E8%AE%BEf%28x%29%3D%28ax%2Bb%29sinx%2B%28cx%2Bd%29cosx+%E8%AF%95%E7%A1%AE%E5%AE%9Aabcd%E7%9A%84%E5%80%BC%2C%E4%BD%BFf%27%28x%29%3Dxsinx-cosx+%E7%AD%94%E6%A1%88%E6%98%AFc%3D1+b%3D-2+a%3Dd%3D0%E8%80%8C%E6%88%91%E6%98%AF+c%3D-1+b%3D2+a%3Dd%3D0)
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设f(x)=(ax+b)sinx+(cx+d)cosx 试确定abcd的值,使f'(x)=xsinx-cosx 答案是c=1 b=-2 a=d=0而我是 c=-1 b=2 a=d=0
设f(x)=(ax+b)sinx+(cx+d)cosx 试确定abcd的值,使f'(x)=xsinx-cosx 答案是c=1 b=-2 a=d=0
而我是 c=-1 b=2 a=d=0
设f(x)=(ax+b)sinx+(cx+d)cosx 试确定abcd的值,使f'(x)=xsinx-cosx 答案是c=1 b=-2 a=d=0而我是 c=-1 b=2 a=d=0
你的结果是正确的,我求的结果也是这个,可能答案错了
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