Ta có: \(D=\frac{19^{209}+1}{19^{210}+1}< \frac{19^{209}+1+18}{19^{210}+1+18}\)
Mà:\(\frac{19^{209}+1+18}{19^{210}+1+18}=\frac{19^{209}+19}{19^{210}+19}=\frac{19\cdot\left(19^{208}+1\right)}{19\cdot\left(2^{209}+1\right)}=\frac{19^{208}+1}{19^{209}+1}=C\)
=> D < C
Cho \(H=\frac{119^{208}+1}{119^{209}+1};K=\frac{119^{209}+1}{119^{210}+1}\)
So sánh H và K
so sanh M va N
M=\(\frac{19^{30}+5}{19^{31}+5}\)
N=\(\frac{19^{31}+5}{19^{32}+5}\)
SO SANH \(\frac{10^{19}+1}{10^{20}+1}\) VA \(\frac{10^{20}+1}{10^{21}+1}\)
A= 1 + 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 +***+2^ 208 +2^ 209 . và B= 1.2 * 0.2 ^ 2 * 0.2 ^ 3 * 0.2 ^ 4 ....2^ 19 .2^ 20 So sánh A và B.
a, \(\frac{19}{18}va\frac{2011}{2010}\)
b, \(\frac{72}{73}va\frac{98}{99}\)
c, \(\frac{7}{9}va\frac{19}{17}\)
11/29+9-17+10-19 va 1/2. Hay so sanh
Tính:
a) \(A=\frac{(1+17)(1+\frac{17}{2})(1+\frac{17}{3})...(1+\frac{17}{19})}{(1+19)(1+\frac{19}{2})(1+\frac{19}{3})...(1+\frac{19}{17})}\)
b) \(B=\frac{1}{-2}.\frac{1}{3}+\frac{1}{-3}.\frac{1}{4}+...+\frac{1}{-5}.\frac{1}{10}\)
c) \(C=(1-\frac{1}{1.2})+(1-\frac{1}{2.3})+...+(1-\frac{1}{2015.2016})\)
d) \(D=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+...+\frac{1}{10}}\)
Giúp mk với:
a,Cho n thuộc N, chứng tỏ rằng : n ( n+1) ( n+2) chia hết cho 6
b, So sánh A và B biết:
A= 19208+1/ 19200+ 1
B= 19200+1/ 19210 +1
So sánh
\(\frac{19^{30}+5}{^{19^{31}}-5}va\frac{19^{31}+5}{19^{32}+5}\)
