5.0 CÂU 4. 3.0A) 3.0 A) A + 2B = C (1), A3 + 8B3 = C2 (2) 2....
Câu 3. 5.0 Câu 4. 3.0
a) 3.0 a) a + 2b = c (1), a
3
+ 8b
3
= c
2
(2) 2.0
0.25
(2) (a + 2b)(a
2
– 2ab + 4b
2
) = c
2
(3)
0.2
= = = = = =
h , h , h
a
2S 2
b
2S 2
c
2S 2
T (1) và (3) suy ra: ừ
a a b b c c
5
(2) a
2
– 2ab + 4b
2
= (a + 2b)
Ta có:MA.GA MB.GB MA.GC
� �
4b
2
– 2(a + 1)b + a
2
– a = 0 (4)
= � + + �
T 2 a.GA b.GB c.GC
’ = (a + 1)
2
– 4(a
2
– a) = –3a
2
+ 6a + 1
3 a.m b.m c.m
(4) có nghi m ệ ’ ≥ 0
a
c
c
3a
2
– 6a 1 3(a – 1)
2
4
a = 1 ho c a = 2 (vì a ặ N*)
2
2
2
2
2
2
2
= + − = + −
a.m a 2b 2c a 3a (2b 2c a )
a
1 1
2 2 3
+ a = 1 b = 1, c = 3
0.25 0.25
+ a = 2 b = 1, c = 4
2
2
2
V y (a;b;c) =(1;1;3) ho c (a;b;c) =(2;1;4) ậ ặ
+ +
a.m 2 3
b) 1.0
a
a b c
n
n 1
1
f (x) a x = + a
−
x
−
+ + ... a x + a
n
n 1
1
0
2
2
2
2
2
2
+ + + +
Gi s : ả ử
b.m , c.m
b
a b c
c
a b c
Ta có: f(a + b) – f(a) =
2 3 2 3
n
n
n 1
n 1
0.2 5
a [(a+b) − a ] a +
−
[(a+b)
−
− a
−
]+...+a b
T 6 3 (MA.GA MB.GB MC.GC) (1)
=
−
−
−
−
n 1
n 2
n 2
n 1
= + + +
a b c + +
a b[(a+b) a(a+b) +...+a (a b) a ]
0,2
n
n 2
n 3
n 3
n 2
+ + +
+a b[(a+b) a(a+b) +...+a (a b) a ]
−
n 1
MA.GA MB.GB MC.GC MA.GA MB.GB MC.GC + + uuuur uuur uuur uuur uuur uuur + +
+...+a b
1