✔ 最佳答案
(1)設-1<=x<=2時,f(x)=x^2-2mx+2m+3之值恆正,求實數m之範圍?
Sol
x^2-2mx+2m+3
=(x-m)^2+(-m^2+2m+3)
頂點(m,-m^2+2m+3)
(1) m>=2
f(2)=4-4m+2m+3>0
7>2m
7/2>m
2<=m<7/2………………….
(2) -1<=m<=2
-m^2+2m+3>0
m^2-2m-3<0
(m-3)(m+1)<0
-1<m<3
-1<m<=2…………………
(3) m<-1
F(-1)=1+2m+2m+3>0
m>-1
無解
綜合(1),(2),(3)
-1<m<7/2
(2)求x^2-5x-2(x^2-5x+3)^(1/2)=0之解為?
Sol
5^2-4*1*3=13>0
x^2-5x+3>0
Set p=√(x^2-5x+3)>0
p^2=x^2-5x+3
x^2-5x-2(x^2-5x+3)^(1/2)=0
(p^2-3)-2p=0
p^2-2p-3=0
(p-3)(p+1)=0
p=3 or p=-1(不合)
x^2-5x+3=9
x^2-5x-6=0
(x-6)(x+1)=0
x=6 or x=-1