Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)
ta có: \(\left\{{}\begin{matrix}\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{1}{3}\\\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\\\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\end{matrix}\right.\)
\(\Rightarrow S< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\Leftrightarrow5S< 4^{\left(1\right)}\)
Lại có: \(\left\{{}\begin{matrix}\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\\\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\\\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{1}{6}\end{matrix}\right.\)
\(\Rightarrow S>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\Leftrightarrow5S>3^{\left(2\right)}\)
từ (1) và (2) => 3<5S<4
A: có 30 số hạng không đủ
phải chia nhỏ ra
\(A=\left(\frac{1}{31}+...+\frac{1}{36}\right)+\left(\frac{1}{37}+..+\frac{1}{48}\right)+\left(\frac{1}{49}+..+\frac{1}{60}\right)\)
\(A>\left(\frac{6}{36}\right)+\left(\frac{12}{48}\right)+\left(\frac{12}{60}\right)=\frac{3}{12}+\frac{3}{12}+\frac{1}{12}=\frac{7}{12}\)
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S=\left[\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right]+\left[\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right]+\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right]\)
\(S< \left[\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right]+\left[\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right]+\left[\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right]\)
\(S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}\)
\(S< \frac{37}{60}< \frac{48}{60}=\frac{4}{5}\)
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{60}\)\(CMR:S< \frac{4}{5}\)
Số số hạng của S là: (60 - 31 ) + 1 = 30 ( số ), chia thành 6 nhóm, mỗi nhóm 5 số hạng.
Ta có:
\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+\frac{1}{35}< \frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}\)
\(\frac{1}{36}+\frac{1}{37}+\frac{1}{38}+\frac{1}{39}+\frac{1}{40}< \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+\frac{1}{45}< \frac{1}{41}+\frac{1}{41}+\frac{1}{41}+\frac{1}{41}+\frac{1}{41}\)
\(\frac{1}{46}+\frac{1}{47}+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}< \frac{1}{46}+\frac{1}{46}+\frac{1}{46}+\frac{1}{46}+\frac{1}{46}\)
\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+\frac{1}{55}< \frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\frac{1}{51}\)
\(\frac{1}{56}+\frac{1}{57}+\frac{1}{58}+\frac{1}{59}+\frac{1}{60}< \frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}\)
\(=>S=\frac{5}{31}+\frac{5}{36}+\frac{5}{41}+\frac{5}{46}+\frac{5}{51}+\frac{5}{56}\)
\(=>S< 0,78...\)\(=>S< \frac{7}{10}\)( mình ước lượng thôi nha )
Vậy \(S< \frac{4}{5}\)vì \(\frac{4}{5}=\frac{8}{10}< \frac{7}{10}\)
~UMK..., mình ko chắc đúng ko nữa~
10=2299495565695652