1.Tính nhanh:
A= \(\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}\)
2. Cho: B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\) .Hãy chứng tỏ rằng B > 1.
3. Rút gọn:
a) C= \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{20}\right)\)
b) D= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}\)
4. So sánh: E=\(\dfrac{20^{10}+1}{20^{10}-1}\) và F =\(\dfrac{20^{10}-1}{20^{10}-3}\)
5. Tính giá trị của biểu thức:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
Chứng minh rằng :
\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
chứng minh rằng :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\) b)\(\dfrac{1}{5^2}+\dfrac{1}{6^5}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
Chứng minh rằng :
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014^2}>\dfrac{1}{5}\)
Cho: S=\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2018^2}\)chứng tỏ S< \(\dfrac{1}{4}\)
chứng minh rằng :
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Chứng tỏ rằng:
a) \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{2}\)
b) \(S=\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{80}>\dfrac{7}{12}\)
c) \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
d) \(\dfrac{49}{100}< S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}< 1\)
Các bạn giải ra từng bước dùm mik nha
Thanks m.n
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)
Chứng minh rằng : \(\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+........+\dfrac{1}{2020^2}< \dfrac{1}{4}\)