CMR:
\(\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+...+\frac{1}{2013^2}< \frac{1}{5}\)
CMR :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2013}-\frac{1}{2014}=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-1-\frac{1}{2}-...-\frac{1}{1007}\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\) (đpcm)
cmr \(s< \frac{1}{3}\)biết \(S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2013}{5^{2013}}+\frac{2014}{5^{2014}}\)
em thử nhân S với 5 rồi lấy 5S= S thử đi
chị làm toàn như vậy
ko bt có đc ko nữa
\(\frac{1}{3^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{9^2}+.....+\frac{1}{2013^2}\)<\(\frac{1}{5}\)
ta có:
1/3^2= 1/9<1/5
1/6^2=1/36<1/5
.
.
.
1/2013^2=1/4052169<1/5
=>1/3^2+1/6^2+....+1/2013^2<1/5
bạn ơi cách bạn ko có hiệu quả đâu
đúng là bt ấy nhỏ hơn \(\frac{1}{5}\)nhưng bạn làm vậy lá sai ngay đấy
thế mình hỏi bạn 1<2;1.1<2nhuwng tổng 1+1.1>2
với phân số cũng thế
Bài 1 : Tính :
a)\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\times230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right)\div\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) \(\frac{2^{12}\times3^5-4^6\times9^2}{\left(2^4\times3\right)^6+8^4\times3^5}-\frac{5^{10}\times7^3-25^5\times49^2}{\left(125\times7\right)^3+5^9\times14^3}\)
c)P=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+....+\frac{1}{2015}}\)
tính
a, \(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
b , \(\left(\frac{0,4-\frac{8}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
c, A
= \(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+...+\frac{1}{20}.\left(1+2+3+....+20\right)\right).155\)
\(a,\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.7}=\frac{2.6}{3.7}=\frac{4}{7}\)
\(M=\hept{\begin{cases}0.4-\frac{2}{9}+\frac{2}{11}-\frac{1}{3}-0.25+\frac{1}{5}\\1,4-\frac{7}{9}+\frac{7}{11}-1\frac{1}{6}-0.875+0.7\end{cases}}\left\{\right\}:\frac{2012}{2013}\)
cái dấu trước dấu chia mình ko biết cách đóng ngoặc nhọn lại nên mình viết như vậy
1) Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
CMR
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{2013^2-1}+\sqrt{2013^2}}=2012\)
Xét số hạng tổng quát: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\) (do \(\sqrt{n+1}-\sqrt{n}>0\forall n\in\mathbb{N}\text{ nên ta có thể nhân liên hợp}\))
Áp dụng vào và ta có:
\(VT=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2013^2}-\sqrt{2013^2-1}\)
\(=\sqrt{2013^2}-1=2013-1=2012^{\left(đpcm\right)}\)