Question

第三题怎么做

怎么做第三题

Answer 1

f(x)=[2(x+3) - 5]/(x+3)
=2(x+3)/(x+3) - 5/(x+3)
=2 - 5/(x+3)
设0<x1<x2
则f(x1) - f(x2)=2 - 5/(x1 + 3) - [2 - 5/(x2 + 3)]
=2 - 5/(x1 + 3) - 2 + 5/(x2 + 3)
=5/(x2 + 3) - 5/(x1 + 3)
=[5(x1 + 3) - 5(x2 + 3)]/(x1 + 3)(x2 + 3)
=(5x1 + 15 - 5x2 - 15)/(x1x2 + 3x1 + 3x2 + 9)
=[5(x1 - x2)]/[x1x2 + 3(x1 + x2) + 9]
∵0<x1<x2
∴x1 - x2<0，x1 + x2>0，x1x2>0
∴5(x1 - x2)<0，
x1x2 + 3(x1 + x2) + 9>0
∴f(x1) - f(x2)<0，即：f(x1)<f(x2)
∴f(x)在(0，+∞)上单调递增

Answer 2

我数学也不好肯定单调递增

Answer 3