Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(B=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
...
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)
Vậy A<\(\frac{3}{4}\)
A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)=\(\frac{2013}{2014}\)<\(\frac{3}{4}\)
bài 1 :
\(\frac{2}{3}\)+\(\frac{1}{3}\)=\(\frac{3}{3}\)=1
\(\frac{3}{4}\)+\(\frac{2}{4}\)+\(\frac{1}{4}\)=\(\frac{4}{4}\)=1
\(\frac{4}{5}\)+\(\frac{3}{5}\)+\(\frac{2}{5}\)+\(\frac{1}{5}\)=\(\frac{10}{5}\)= 2
chúc bạn học tốt !!!
Trả lời
a)
\(x^2:\frac{16}{11}=\frac{11}{4}\)
\(\Leftrightarrow x^2=\frac{11}{4}\cdot\frac{16}{11}\)
\(\Leftrightarrow x^2=\frac{16}{4}\)
\(\Leftrightarrow x^2=\left(\frac{4}{2}\right)^2\)
\(\Leftrightarrow x=\frac{4}{2}\)
Vậy x=\(\frac{4}{2}\)
b) (bạn thiếu nhóm \(\frac{1}{10\cdot13}\))
Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\)
\(\Rightarrow3A=3\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\right)\)
\(\Rightarrow3A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\)
\(\Rightarrow3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}\)
\(\Rightarrow3A=1-\frac{1}{19}\Leftrightarrow3A=\frac{18}{19}\)
\(\Rightarrow A=\frac{18}{19}:3\Leftrightarrow A=\frac{6}{19}\)
Câu 3 : \(2+4+6+.........+2n=156\)
\(\Leftrightarrow2\left(1+2+3+.....+n\right)=156\)
\(\Leftrightarrow1+2+3+.........+n=78\)
\(\Leftrightarrow\frac{n\left(n+1\right)}{2}=78\)\(\Leftrightarrow n\left(n+1\right)=156=12.13\)\(\Leftrightarrow n=12\)
Vậy \(n=12\)
\(\frac{-3}{14}\) - \(\frac{5}{-14}\)
= \(\frac{-3}{14}+\frac{5}{14}\)
= \(\frac{2}{14}\)
= \(\frac17\)
- \(\frac54\) - \(\frac34\)
= -(\(\frac54+\frac34\))
= - \(\frac84\)
= - 2
\(\frac{15}{6}-\frac{-10}{20}\)
= \(\frac52\) + \(\frac12\)
= \(\frac62\)
= 3
\(\frac{26}{-35}\) - \(\frac{6}{35}\)
= - (\(\frac{26}{35}\) + \(\frac{6}{35}\))
=- \(\frac{32}{35}\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2018}{2017}.\frac{2019}{2018}\)
\(=\frac{3.4.5.6.....2018.2019}{2.3.4.5....2017.2018}=\frac{2019}{2}\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2019}{2018}\)
\(=\frac{3.4.5....2019}{2.3.4....2018}\)
\(=\frac{2019}{2}\)
Tính :
a) \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2019}{2018}\)
\(=\frac{3.4.5.....2019}{2.3.4.....2018}\)
\(=\frac{2019}{2}\)