CTR: 1/2+1/3+1/4+...+1/100
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=1/2+1/3+1/4+...+1/100
xét mẫu:có ssh là (100-2):1+1=99 số
tổng là (100+2)*99:2=5940
vậy ta có 1/5940
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
a) \(...=-\dfrac{1}{4}.\dfrac{4}{17}.\left(-\dfrac{63}{21}\right).\left(-\dfrac{7}{12}\right)\)
\(=-\dfrac{1}{17}.\dfrac{63}{21}.\dfrac{7}{12}\)
\(=-\dfrac{7}{68}\)
b) \(...=-\dfrac{2}{5}.\dfrac{4}{15}-\dfrac{3}{10}.\dfrac{4}{15}\)
\(=\dfrac{4}{15}\left(-\dfrac{2}{5}-\dfrac{3}{10}\right)\)
\(=\dfrac{4}{15}\left(-\dfrac{4}{10}-\dfrac{3}{10}\right)\)
\(=\dfrac{4}{15}.\left(-\dfrac{7}{10}\right)=-\dfrac{14}{75}\)
c) \(...=21-\dfrac{15}{4}:\left(\dfrac{9}{24}-\dfrac{4}{24}\right)\)
\(=21-\dfrac{15}{4}:\dfrac{5}{24}\)
\(=21-\dfrac{15}{4}.\dfrac{24}{5}\)
\(=21-3.6=3\)
d) \(...=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right).\dfrac{7}{3}\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right)\)
\(=\dfrac{7}{3}\left(-1+1\right)=0\)
\(-\dfrac{1}{2}=\dfrac{-1\times5}{2\times5}=-\dfrac{5}{10}\\ -\dfrac{1}{3}=\dfrac{-1\times5}{3\times5}=-\dfrac{5}{15}\\ -\dfrac{5}{10}>-\dfrac{5}{11};-\dfrac{5}{12};-\dfrac{5}{13};-\dfrac{5}{14}>-\dfrac{5}{15}\\ \Rightarrow a\in\left\{-\dfrac{5}{11};-\dfrac{5}{12};-\dfrac{5}{13};-\dfrac{5}{14}\right\}\)
Câu a, xem lại đề bài
Câu b:
P = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ...+ \(\dfrac{1}{2023^2}\)
Vì \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{4^2}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)
........................
\(\dfrac{1}{2023^2}\) < \(\dfrac{1}{2022.2023}\) = \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\)
Cộng vế với vế ta có:
0< P < 1 - \(\dfrac{1}{2023}\) < 1
Vậy 0 < P < 1 nên P không phải là số tự nhiên vì không tồn tại số tự nhiên giữa hai số tự nhiên liên tiếp
Câu c:
C = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{2021^2}\) + \(\dfrac{1}{2023^2}\) = C
B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+ \(\dfrac{1}{2020^2}\) + \(\dfrac{1}{2023^2}\) > 0
Cộng vế với vế ta có:
C+B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\)+ \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{2023^2}\) > C + 0 = C > 0
Mặt khác ta có:
1 > \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{2023^2}\) (cm ở ý b)
Vậy 1 > C > 0 hay C không phải là số tự nhiên (đpcm)
https://olm.vn/cau-hoi/a-cho-a12211216211002-ctr-a12-b-cho-p122132142120232-ctr-p-khong-la-so-tu-nhien-c-cho-c132152172120211.8293222842881
Cô làm rồi em nhá
\(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{100}\)
\(\Rightarrow A=\dfrac{99}{100}\)
Đoạn suy ra đầu tiên cơ sở gì bạn suy ra được như vậy nhỉ?