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证明:函数y=x+1/x在(0,1)上是减函数
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设0<x1<x2<1
f(x2)-f(x1)
=x2+1/x2-x1-1/x1
=(x2-x1)(1-1/x2x1)
x2<x1∴x2-x1<0
0<x2<x1<1
∴0<x2x1<1
1/x2x1>1
1-1/x2x1<0
∴f(x2)-f(x1)<0
函数y=x+1/x在(0,1)上是减函数</x2x1<1
</x2<x1<1
</x1∴x2-x1<0
</x1<x2<1
f(x2)-f(x1)
=x2+1/x2-x1-1/x1
=(x2-x1)(1-1/x2x1)
x2<x1∴x2-x1<0
0<x2<x1<1
∴0<x2x1<1
1/x2x1>1
1-1/x2x1<0
∴f(x2)-f(x1)<0
函数y=x+1/x在(0,1)上是减函数</x2x1<1
</x2<x1<1
</x1∴x2-x1<0
</x1<x2<1
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