Question

设x1,x2为方程4x2-4mx+m+2的两个根,求x12+x22的最小值？

Answer 1

有根,判别式大于等于0
16m²-16(m+2)>=0
m²-m-2=(m+1)(m-2)>=0
m=2
x1+x2=m
x1x2=(m+2)/4
x1²+x2²
=(x1+x2)²-2x1x2
=m²-2(m+2)/4
=(2m²-m-2)/2
=[2(m-1/4)²-17/8]/2
m=2
所以m=-1,最小值=1/2,2,把f(x)g(x)看做整体
则=[f(x)g(x)]'*h(x)+f(x)g(x)*h'(x)
=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)
[f(x)/g(x)/h(x)]'
也是把f(x)/g(x)看做整体来解决 把f(x)g(x)看做整体
则=[f(x)g(x)]'*h(x)+f(x)g(x)*h'(x)
,1,