VN
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DY
7
S
3 tháng 5 2019
A = 3/4 + 8/9 + 15/16 + ... + 399/400
A = 1 - 1/4 + 1 - 1/9 + 1 - 1/16 + ... + 1 - 1/400
A = (1 + 1 + 1 + ... +1) - (1/4 + 1/9 + 1/16 + ... + 1/400)
A = 19 - (1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20)
đặt b = 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20
có 1/2.2 < 1/1.2 ; 1/3.3 < 1/2.3 ; ... 1/20.20 < 1/19.20
=> b < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/19.20
=> b < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/19 - 1/20
=> b < 1 - 1/20
=> b < 1
mà A = 19 - b
=> A > 18
3 tháng 5 2019
\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)
\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+.....+\frac{20^2-1}{20^2}\)
\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{20^2}\right)\)
\(>19-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{19\cdot20}\right)\)
\(=19-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{19}-\frac{1}{20}\right)\)
\(=19-\left(1-\frac{1}{20}\right)\)
\(>19-1=18\)
LH
1
16 tháng 2 2022
\(8x-48+4x-12-14=-x+4\)
\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)
\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)
\(\Leftrightarrow2x=130\Leftrightarrow x=65\)
\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)
\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)
\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)
25 tháng 2 2019
\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)
\(\Rightarrow M=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+....+\frac{20^2-1}{20^2}\)
\(\Rightarrow M=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+\frac{4^2}{4^2}-\frac{1}{4^2}+....+\frac{20^2}{20^2}-\frac{1}{20^2}\)
\(\Rightarrow M=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{20^2}\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{20^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{19\cdot20}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{19}{20}< 1\)
\(\Rightarrow A< 1\)
\(\Rightarrow M>18\)
8 tháng 5 2016
A=(1-\(\frac{1}{4}\))+(1-\(\frac{1}{9}\))+(1-\(\frac{1}{16}\))+...+(1-\(\frac{1}{400}\)).
A=19-(\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\))
Ta thấy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}<1\)
=>A>19-1=18(đpcm)
CD
1
6 tháng 5 2016
Xét A= \(\frac{3}{4}\)+ \(\frac{8}{9}\) +...+ \(\frac{399}{400}\)
= (1 - \(\frac{1}{2^2}\)) + (1- \(\frac{1}{3^2}\)) +...+ (1- \(\frac{1}{20^2}\))
= (1+1+1+...+1) - (\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\)) Bạn phải mở ngoặc có 19 số 1 nha!
= 19 - (\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\))
Đặt B =\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\) < \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +...+ \(\frac{1}{19.20}\) = 1- \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) +...+ \(\frac{1}{19}\) - \(\frac{1}{20}\) = 1 - \(\frac{1}{20}\) = \(\frac{19}{20}\)
=> A= 19 - B= 18+ 1- \(\frac{19}{20}\) >18 => A>18
14 tháng 10 2018
\(a)\)\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)
\(M=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{400-1}{400}\)
\(M=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{400}\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{400}\right)\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)
Do từ 2 đến 20 có \(20-2+1=19\) nên :
\(M=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\)
\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)
\(A>\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow\)\(M=19-A>19-\frac{1}{2}+\frac{1}{21}=18,5+\frac{1}{21}>8\)
\(\Rightarrow\)\(M>8\) ( đpcm )
Còn câu b) bn xem lại đề đi, nếu đề đúng thì mk sai :v
Chúc bạn học tốt ~
21 tháng 1 2019
\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}...+\frac{399}{400}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+\left(1-\frac{1}{25}\right)+...+\left(1-\frac{1}{400}\right)\)
\(=\left(1+1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{20^2}\right)\)
\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)
Đặt \(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{20^2}\)
\(< P=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{20\cdot21}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\)
\(=\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow M+N>19-\frac{1}{2}+\frac{1}{21}=\frac{37}{2}+\frac{1}{21}>8\)
b sai đề.chừng nào chữa đề thì làm
\(\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot...\cdot\dfrac{400}{399}\)
\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot...\cdot\dfrac{20\cdot20}{19\cdot21}\)
\(=\dfrac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot20\cdot20}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot19\cdot21}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot20}{1\cdot2\cdot3\cdot...\cdot19}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot20}{3\cdot4\cdot5\cdot...\cdot21}\)
\(=20\cdot\dfrac{2}{21}\)
\(=\dfrac{40}{21}\)