已知f(x)=bx+1/2x+a a,b为常数,且ab不等于2 f(x)f(1/x)=k,求K的值解:f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax) k=f(x)f(1/x)=[(bx+1)/(2x+a)][(b+x)/(2+ax)] =(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)] x+1/b=x+2/a且b+x=x+a/2 中最后一步怎么求(b/2
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![已知f(x)=bx+1/2x+a a,b为常数,且ab不等于2 f(x)f(1/x)=k,求K的值解:f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax) k=f(x)f(1/x)=[(bx+1)/(2x+a)][(b+x)/(2+ax)] =(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)] x+1/b=x+2/a且b+x=x+a/2 中最后一步怎么求(b/2](/uploads/image/z/573348-12-8.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dbx%2B1%2F2x%2Ba%E3%80%80a%2Cb%E4%B8%BA%E5%B8%B8%E6%95%B0%2C%E4%B8%94ab%E4%B8%8D%E7%AD%89%E4%BA%8E2+f%28x%29f%281%2Fx%29%3Dk%2C%E6%B1%82K%E7%9A%84%E5%80%BC%E8%A7%A3%3Af%281%2Fx%29%3D%28b%2Fx%2B1%29%2F%282%2Fx%2Ba%29%3D%28b%2Bx%29%2F%282%2Bax%29+k%3Df%28x%29f%281%2Fx%29%3D%5B%28bx%2B1%29%2F%282x%2Ba%29%5D%5B%28b%2Bx%29%2F%282%2Bax%29%5D+%3D%28b%2F2a%29%5B%28x%2B1%2Fb%29%2F%28x%2Ba%2F2%29%5D%5B%28b%2Bx%29%2F%28x%2B2%2Fa%29%5D+x%2B1%2Fb%3Dx%2B2%2Fa%E4%B8%94b%2Bx%3Dx%2Ba%2F2+%E4%B8%AD%E6%9C%80%E5%90%8E%E4%B8%80%E6%AD%A5%E6%80%8E%E4%B9%88%E6%B1%82%28b%2F2)
已知f(x)=bx+1/2x+a a,b为常数,且ab不等于2 f(x)f(1/x)=k,求K的值解:f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax) k=f(x)f(1/x)=[(bx+1)/(2x+a)][(b+x)/(2+ax)] =(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)] x+1/b=x+2/a且b+x=x+a/2 中最后一步怎么求(b/2
已知f(x)=bx+1/2x+a a,b为常数,且ab不等于2 f(x)f(1/x)=k,求K的值
解:f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax)
k=f(x)f(1/x)=[(bx+1)/(2x+a)][(b+x)/(2+ax)]
=(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)]
x+1/b=x+2/a且b+x=x+a/2 中最后一步怎么求
(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)] 然后就知道
x+1/b=x+2/a且b+x=x+a/2这一步怎么得来的
已知f(x)=bx+1/2x+a a,b为常数,且ab不等于2 f(x)f(1/x)=k,求K的值解:f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax) k=f(x)f(1/x)=[(bx+1)/(2x+a)][(b+x)/(2+ax)] =(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)] x+1/b=x+2/a且b+x=x+a/2 中最后一步怎么求(b/2
解题步骤是这样的:解:f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax)
k=f(x)f(1/x)=[(bx+1)/(2x+a)][(b+x)/(2+ax)]
=(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)]
x+1/b=x+2/a且b+x=x+a/2
∴1/b=2/a且b=a/2
a=2b
k=(b/4b)[(x+1/b)/(x+b)][(b+x)/(x+1/b)]
=1/4
∴k=1/4
但是[(bx+1)/(2x+a)][(b+x)/(2+ax)]
=(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)]
1+1=2
最后一步不就是K=b/2a(a,b为常数)吗?如果按照你的过程
解题步骤是这样的:解:f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax)
k=f(x)f(1/x)=[(bx+1)/(2x+a)][(b+x)/(2+ax)]
=(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)]
x+1/b=x+2/a且b+x=x+a/2
∴1/b=2/a且b=a/2
a=2b
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解题步骤是这样的:解:f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax)
k=f(x)f(1/x)=[(bx+1)/(2x+a)][(b+x)/(2+ax)]
=(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)]
x+1/b=x+2/a且b+x=x+a/2
∴1/b=2/a且b=a/2
a=2b
k=(b/4b)[(x+1/b)/(x+b)][(b+x)/(x+1/b)]
=1/4
∴k=1/4
但是[(bx+1)/(2x+a)][(b+x)/(2+ax)]
=(b/2a)[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)]
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