Đặt \(S=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}\)
Ta thấy :
\(\frac{1}{201}>\frac{1}{400}\)
\(\frac{1}{202}>\frac{1}{400}\)
...
\(\frac{1}{399}>\frac{1}{400}\)
\(\Rightarrow S>\frac{1}{400}+\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\)
có 200 dãy \(\Rightarrow S>\frac{200}{400}=\frac{1}{2}\)
Vậy : \(S>\frac{1}{2}\)
CMR : \(\frac{1}{201}\) + \(\frac{1}{202}\) + ... + \(\frac{1}{399}\) + \(\frac{1}{400}\) > \(\frac{1}{2}\)
Chứng tỏ :
S = \(\frac{1}{201}\)+ \(\frac{1}{202}\)+.........+\(\frac{1}{399}\)+\(\frac{1}{400}\)>\(\frac{1}{2}\)
A = \(\frac{10}{27}\)+ \(\frac{9}{16}\)+ \(\frac{11}{34}\)< 2
chứng tỏ rằng :
\(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+....+\frac{1}{400}>\frac{1}{2}\)
so sánh:
\(\frac{2009}{2010}và\frac{2010}{2011}\)\(\frac{1}{3^{400}}và\frac{1}{4^{300}}\)
\(\frac{200}{201}+\frac{201}{202}và\frac{200+201}{201+202}\)\(\frac{2008}{2008.2009}và\frac{2009}{2009.2010}\)
Cho \(A=\frac{1}{2}+\frac{1}{301}+\frac{1}{302}+\frac{1}{303}+...+\frac{1}{399}+\frac{1}{400}\)
CMR: A<1
Giải chi tiết giùm ak!
So sánh phân số
A/ \(\frac{2009}{2010}\)và\(\frac{2010}{2011}\)
B/ \(\frac{1}{3^{400}}\) và \(\frac{1}{4^{300}}\)
C/\(\frac{200}{201}+\frac{201}{202}và\frac{200+201}{201+202}\)
D/\(\frac{2008}{2008\cdot2009}và\frac{2009}{2009\cdot2010}\)
1/TÍNH NHANH
a/ \(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
2/so sánh
a/\(\frac{2009}{2010}va\frac{2010}{2011}\) b/\(\frac{1}{3^{400}}va\frac{1}{4^{300}}\) c/\(\frac{200}{201}+\frac{201}{202}va\frac{200+201}{201+202}\) d/\(\frac{2008}{2008+2009}va\frac{2009}{2009+2010}\)
3/TÌM X BIẾT
\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\right)-x=\frac{-100}{99}\)
GIÚP MÌNH NHA MAI MÌNH NỘP RÙI
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{Xx\left(X+1\right):2}=\frac{399}{400}\)
1/ So sánh
a/\(\frac{2009}{2010}\)và \(\frac{2010}{2011}\) b/\(\frac{1}{3^{400}}\)và \(\frac{1}{4^{300}}\)
c/\(\frac{200}{201}\)+ \(\frac{201}{202}\)và \(\frac{200+201}{201+202}\) d/\(\frac{2008}{2008.2009}\)và \(\frac{2009}{2009.2010}\)
