Question

在算式4*[]+1*()=30的两个[],()中分别填入两个正整数 使它们倒数之和最小则这两个数[],()应为--

Answer 1

设[]=x,则()=30-4x

y=1/x+1/(30-4x)=(30-4x+x)/x(30-4x)=(3x-30)/(4x^2-30x)

1/y=(4x^2-30x)/(3x-30)

=(4x^2-40x+10x)/(3x-30)

=4x/3+10x/(3x-30)

=4x/3+10/3+100/(3x-30)

=(1/3)[4(x-10)+100/(x-10)]+50/3

>=1/3*2根号(4*100)+50/3

=40/3+50/3

=30

当4(x-10)=100/(x-10)取等号

(x-10)^2=25

x-10=5,-5

x=15,5

x=15,()=30-4x