已知f(x)=(bx+1)/(2x+a),a,b为常数,且a*b≠2,若f(x)·f(1/x)=k,求常数k的值.解题步骤是这样的: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+
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![已知f(x)=(bx+1)/(2x+a),a,b为常数,且a*b≠2,若f(x)·f(1/x)=k,求常数k的值.解题步骤是这样的: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+](/uploads/image/z/13158278-62-8.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3D%28bx%2B1%29%2F%282x%2Ba%29%2Ca%2Cb%E4%B8%BA%E5%B8%B8%E6%95%B0%2C%E4%B8%94a%2Ab%E2%89%A02%2C%E8%8B%A5f%28x%29%C2%B7f%281%2Fx%29%3Dk%2C%E6%B1%82%E5%B8%B8%E6%95%B0k%E7%9A%84%E5%80%BC.%E8%A7%A3%E9%A2%98%E6%AD%A5%E9%AA%A4%E6%98%AF%E8%BF%99%E6%A0%B7%E7%9A%84%EF%BC%9Af%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%2B)
已知f(x)=(bx+1)/(2x+a),a,b为常数,且a*b≠2,若f(x)·f(1/x)=k,求常数k的值.解题步骤是这样的: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+
已知f(x)=(bx+1)/(2x+a),a,b为常数,且a*b≠2,若f(x)·f(1/x)=k,求常数k的值.
解题步骤是这样的: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x+1/b=x+2/a且b+x=x+a/2 这一步我不明白,
已知f(x)=(bx+1)/(2x+a),a,b为常数,且a*b≠2,若f(x)·f(1/x)=k,求常数k的值.解题步骤是这样的: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+
因为K为常数那么[(x+1/b)/(x+a/2)][(b+x)/(x+2/a)] 中的分子分母一定要可相互约分
所以(x+1/b)/(x+a/2)中的分子要和第二个因式的分母约分,它的分母要和第二个式子中的分子约分,否则f(x)·f(1/x)=k就不会是常数