\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=20.\frac{1}{60}=\frac{1}{3}\)
=> \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}>\frac{1}{3}\)
CM::
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+............+\frac{1}{79}+\frac{1}{80}>\frac{7}{12}\)
Chứng minh
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+........................+\frac{1}{78}+\frac{1}{79}+\frac{1}{80}< \frac{1}{2}\)
Giúp mình nhé ai nhanh nhất mình tick
1) có thể tìm đc 2 số nguyên x,y sao cho 45x+10y=-20152016 ko?
2)CMR:A=\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+\frac{1}{45}+...+\frac{1}{80}>\frac{7}{12}\)
chứng tỏ rằng :\(\frac{1}{8}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{41}+\frac{1}{42}+\frac{1}{43}< \frac{1}{2}\)
Chứng tỏ rằng: \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}>\frac{7}{12}\)
CHỨNG TỎ
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+......+\frac{1}{99}+\frac{1}{100}>\frac{7}{10}\)
So sánh M = \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\) và \(\frac{4}{5}\)
HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Chứng tỏ rằng:
a/ \(\frac{1}{2}< \frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}< 1\)
b/ \(1< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< 2\)
c/ A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}< 1\)
d/ \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}< \frac{1}{2}\)
e/ \(\frac{2}{5}< \frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}< \frac{2}{3}\)
f/\(C=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{19}{9^2\cdot10^2}< 1\)
Cm: \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}>\frac{7}{12}\)
