K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

3 tháng 4 2018

\(B=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(\Leftrightarrow B=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(\Leftrightarrow B=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)

\(\dfrac{3}{8}< \dfrac{1}{2}\)

\(\Rightarrow B< \dfrac{1}{2}\left(ĐPCM\right)\)

27 tháng 7 2021

\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)

\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(M=2.\dfrac{3}{16}\)

\(M=\dfrac{3}{8}\)

Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)

3 tháng 2 2023

 

b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử  \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(B\text{=}2.63+...+2^{56}.63\)

\(\Rightarrow B⋮63\)

\(\Rightarrow B⋮21\)

 

17 tháng 3 2017

b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)

\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)

\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)

\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)

\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(2x=93-3=90\)

\(\Rightarrow x=90:2=45\)

19 tháng 3 2017

Cảm ơn bạn

2 tháng 5 2023

a) Ta có \(A=\dfrac{n-5}{n-3}=\dfrac{n-3-2}{n-3}=1-\dfrac{2}{n-3}\). Để \(A\inℤ\) thì \(\dfrac{2}{n-3}\inℤ\) hay \(n-3\) là ước của 2. Suy ra \(n-3\in\left\{\pm1;\pm2\right\}\)

Nếu \(n-3=1\Rightarrow n=4\)\(n-3=-1\Rightarrow n=2\)\(n-3=2\Rightarrow n=5\)\(n-3=-2\Rightarrow n=1\). Vậy để \(A\inℤ\) thì \(n\in\left\{1;2;4;5\right\}\)

 \(A=\dfrac{n+4}{n+1}\) làm tương tự.

b) Dễ thấy các số ở mẫu có thể viết dưới dạng:

\(10=1+2+3+4=\dfrac{4\left(4+1\right)}{2}=\dfrac{4.5}{2}\)

\(15=1+2+3+4+5=\dfrac{5\left(5+1\right)}{2}=\dfrac{5.6}{2}\)

\(21=1+2+...+6=\dfrac{6\left(6+1\right)}{2}=\dfrac{6.7}{2}\)

...

\(120=1+2+...+15=\dfrac{15\left(15+1\right)}{2}=\dfrac{15.16}{2}\)

Do đó \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\) 

\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(A=\dfrac{3}{8}\)

 

15 tháng 4 2017

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...\dfrac{1}{240}\right)\)

\(=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+...+\dfrac{1}{15\times16}\right)\)

\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(=\dfrac{3}{8}\)

15 tháng 4 2017

=2/20+2/30+2/42+.....+2/240

=2/4.5+2/5.6+2/6.7+.....+2/15.16

=1/2[1/4.5+1/5.6+1/6.7+.....+1/15.16]

=1.2[1/4-1/5+1/5-1/6+.....+1/15-1/16]

=1/2[1/4-1/16]

=1/2.3/16

=3/32

12 tháng 7 2017

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)

\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)

\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)

10 tháng 4 2018

A= \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\)

= 1-\(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{6}\)

= 1 - \(\dfrac{1}{6}\)= \(\dfrac{5}{6}\)

mk chỉ bt làm câu 1 thôi ak

mong bn thông cảmthanghoa

2 tháng 7 2017

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{240}\right)\)

\(A=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+....+\dfrac{1}{15\times16}\right)\)

\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)

2 tháng 7 2017

b) cậu đi tìm số sốm hạng là : \(\left(2010-1\right):1+1=2010\)

\(\Rightarrow\)số cặp trong phép tính là : \(2010:2=1005\)(cặp)

\(\Rightarrow B=1-2+3-4+...+2009-2010\)(1005 cặp)

\(\Rightarrow\left(1-2\right)+\left(3-4\right)+...+\left(2009-2010\right)\)

\(\Rightarrow B=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)(1005 số -1)

\(\Rightarrow B=\left(-1\right).1005\)

\(\Rightarrow B=\left(-1005\right)\)

cậu tik cho mik nhé!!!

11 tháng 3 2017

\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)

\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(A=2.\dfrac{3}{16}\)

\(A=\dfrac{3}{8}\)

11 tháng 3 2017

\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)

\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(B=\dfrac{1}{3}-\dfrac{1}{111}\)

\(B=\dfrac{12}{37}\)