设P(x,y)为函数y=x-1(x>√3)图像上一动点,记m=(3x+y-5)/(x-1)+(x+3y-7)/(y-2)

问题描述:

设P(x,y)为函数y=x-1(x>√3)图像上一动点,记m=(3x+y-5)/(x-1)+(x+3y-7)/(y-2),则当m最小时,点P的坐标是
1个回答 分类:数学 2014-09-19

问题解答:

我来补答
答:
y=x^2-1
m=(3x+y-5)/(x-1)+(x+3y-7)/(y-2)
=(3x+x^2-1-5)/(x-1)+(x+3x^2-3-7)/(x^2-1-2)
=(x^2+3x-6)/(x-1)+(3x^2+x-10)/(x^2-3)
m'(x)=(2x+3)/(x-1)-(x^2+3x-6)/(x-1)^2+(6x+1)/(x^2-3)-(6x^3+2x^2-20x)/(x^2-3)^2
=(x^2-2x+3)[1/(x-1)^2-1/(x^2-3)^2]
=(x^2-2x+3)(x^2+x-4)(x^2-x-2)/[(x-1)^2*(x^2-3)^2]
因为x>√3,因此只有x^2-x-2的值可能为0
令m‘(x)=0,则:x^2-x-2=0,解得:x=2(x=-1不符合舍去)
所以:
当√3
 
 
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