.已知函数f(x)、g(x)在R上有定义,且f(x-y)=f(x)g(y)-g(x)f(y),若f⑴=f⑵≠0,则g⑴+g(-1)= ____
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![.已知函数f(x)、g(x)在R上有定义,且f(x-y)=f(x)g(y)-g(x)f(y),若f⑴=f⑵≠0,则g⑴+g(-1)= ____](/uploads/image/z/13794961-49-1.jpg?t=.%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E3%80%81g%28x%29%E5%9C%A8R%E4%B8%8A%E6%9C%89%E5%AE%9A%E4%B9%89%2C%E4%B8%94f%28x%EF%BC%8Dy%29%EF%BC%9Df%28x%29g%28y%29%EF%BC%8Dg%28x%29f%28y%29%2C%E8%8B%A5f%E2%91%B4%EF%BC%9Df%E2%91%B5%E2%89%A00%2C%E5%88%99g%E2%91%B4%EF%BC%8Bg%28%EF%BC%8D1%29%EF%BC%9D+____)
.已知函数f(x)、g(x)在R上有定义,且f(x-y)=f(x)g(y)-g(x)f(y),若f⑴=f⑵≠0,则g⑴+g(-1)= ____
.已知函数f(x)、g(x)在R上有定义,且f(x-y)=f(x)g(y)-g(x)f(y),若f⑴=f⑵≠0,则g⑴+g(-1)= ____
.已知函数f(x)、g(x)在R上有定义,且f(x-y)=f(x)g(y)-g(x)f(y),若f⑴=f⑵≠0,则g⑴+g(-1)= ____
特殊值法:
令x = 2,y = 1
f(2-1) = f(1) = f(2)g(1) - g(2)f(1)
令x = 1,y = 2
f(1-2) = f(-1) = f(1)g(2) - g(1)f(2) = -f(1)
即
f(-1) = -f(1) .(1)
令x = 1,y = -1
f(1+1) = f(2) = f(1)g(-1) - f(-1)g(1)
代入(1)得
f(1)g(-1) + f(1)g(1) = f(2) = f(1)
两边同除f(1)
g(1) + g(-1) = 1
已知函数f(x)、g(x)在R上有定义,且f(x-y)=f(x)g(y)-g(x)f(y),若f⑴=f⑵≠0,则g⑴+g(-1)= ________.
令x=y=0
∴f(0)=f(0)g(0)-g(0)f(0)
∴f(0)=0
令x=1,y=0
∴f(1)=f(1-0)=f(1)g(0)-g(1)f(0)=f(1)g(0)
∴g(0)=1
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已知函数f(x)、g(x)在R上有定义,且f(x-y)=f(x)g(y)-g(x)f(y),若f⑴=f⑵≠0,则g⑴+g(-1)= ________.
令x=y=0
∴f(0)=f(0)g(0)-g(0)f(0)
∴f(0)=0
令x=1,y=0
∴f(1)=f(1-0)=f(1)g(0)-g(1)f(0)=f(1)g(0)
∴g(0)=1
令x=0,y=1
∴f(-1)=f(0-1)=f(0)g(1)-g(0)f(1)=-g(0)f(1)=-f(1)
令x=-1,y=1
∴f(-2)=f(-1-1)=f(-1)g(1)-g(-1)f(1)=-f(1)g(1)-g(-1)f(1)=-f(1)[g(1)+g(-1)]
又∵f(-2)=f(1)≠0
∴g(1)+g(-1)=-1
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