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证明如下:设0<x1<x2<+∞,
则有f(x2)−f(x1)=x2−
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x2 |
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x1 |
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x1 |
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x2 |
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x2 |
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x1 |
=(x2−x1)+(
x2−x1 |
x1•x2 |
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x1•x2 |
x1x2+1 |
x1•x2 |
1+x1x2 |
x1x2 |
∵0<x1<x2<+∞,x2-x1>0且x1x2+1>0,x1x2>0,
所以f(x2)-f(x1)>0,即f(x1)<f(x2).
所以函数y=f(x)在区间(0,+∞)上单调递增.
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x |
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x2 |
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x1 |
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x1 |
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x2 |
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x2 |
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x1 |
x2−x1 |
x1•x2 |
1 |
x1•x2 |
x1x2+1 |
x1•x2 |
1+x1x2 |
x1x2 |