請問:(1)設-1<=x<=2時,f(x)=x^2-2mx+2m+3之值恆正,求實數m之範圍?(2)求x^2-5x-2[x^-5x+3]^(1/2)=0之解為?謝謝~?

2015-11-29 7:48 am

回答 (2)

2015-11-29 1:09 pm
✔ 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
2015-11-29 9:07 am
1.值恆正表領導係數大於0而判別式小於0
D=(-2m)^2-4(2m+3)<0
==>(m-3)(m+1)<0
==>-1<m<3

2.
先決條件x^2-5x+3>=0
==>x>=(5+根號13)/2或x<=(5-根號13)/2
令y=x^2-5x
原式=y-2根號(y+3)=0
==>(y-6)(y+2)=0
==>y=6或y=-2
x^2-5x=6
==>(x-6)(x+1)=0
==>x=6,-1
x^2-5x=-2
==>x=(5+-根號17)/2


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