0Ủ) TA CÚ CXX+ CXX−1+ CXX−1+ CXX−2 = C2X+X2−3⇔ CXX+1+ CXX+−11= CX2+X2−3⇔ CXX+2 = CX2+X2−3⇔ − = ⇔ =(5 X )
1.0
Cõu
∈
ðK : 2 x 5
VIIb
x N
(1.0ủ)
Ta cú C
x
x
+ C
x
x
−
1
+ C
x
x
−
1
+ C
x
x
−
2
= C
2
x
+
x
2
−
3
⇔ C
x
x
+
1
+ C
x
x
+
−
1
1
= C
x
2
+
x
2
−
3
⇔ C
x
x
+
2
= C
x
2
+
x
2
−
3