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OY
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 \(\dfrac{1}{3}< M< \dfrac{1}{2}\)
Q
16 tháng 4 2017
\(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-\dfrac{2}{20}-\dfrac{2}{30}-\dfrac{2}{42}-...-\dfrac{2}{240}=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\right)=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-2.\dfrac{3}{16}=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-\dfrac{3}{8}=\dfrac{5}{8}\)
\(\dfrac{x}{2008}=\dfrac{5}{8}+\dfrac{3}{8}\)
\(\dfrac{x}{2008}=1=\dfrac{2008}{2008}\)
\(\Rightarrow x=2008\)
3 tháng 4 2018
\(B=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(\Leftrightarrow B=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(\Leftrightarrow B=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(\Leftrightarrow B=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)
Vì \(\dfrac{3}{8}< \dfrac{1}{2}\)
\(\Rightarrow B< \dfrac{1}{2}\left(ĐPCM\right)\)
17 tháng 3 2017
b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
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 \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=\dfrac{3}{8}\)
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
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{111}\)
\(B=\dfrac{12}{37}\)
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 cảm
27 tháng 9 2021
Ta có:
\(\dfrac{1}{3}\times\dfrac{12}{12}=\dfrac{12}{36};\)
\(\dfrac{1}{6}\times\dfrac{6}{6}=\dfrac{6}{36};\)
\(\dfrac{1}{10}\times\dfrac{3}{3}=\dfrac{3}{30};\)
\(\dfrac{1}{15}\times\dfrac{2}{2}=\dfrac{2}{30};\)
\(\dfrac{1}{21}\times\dfrac{4}{4}=\dfrac{4}{84};\)
\(\dfrac{1}{28}\times\dfrac{3}{3}=\dfrac{3}{84};\)
\(A=\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{3}{30}+\dfrac{2}{30}+\dfrac{4}{84}+\dfrac{3}{84}+\dfrac{1}{36}\)
\(=\left(\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{1}{36}\right)+\left(\dfrac{3}{30}+\dfrac{2}{30}\right)+\left(\dfrac{4}{84}+\dfrac{3}{84}\right)\)
\(=\dfrac{19}{36}+\dfrac{5}{30}+\dfrac{7}{84}\)
\(=\dfrac{19}{36}+\dfrac{1}{6}+\dfrac{1}{12}\)
\(=\dfrac{19}{36}+\dfrac{6}{36}+\dfrac{3}{36}\)
\(=\dfrac{28}{36}=\dfrac{7}{9}\)
Vậy: \(A=\dfrac{7}{9}\)
LH
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!!!
\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...\dfrac{1}{240}\right)\)
\(=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+...+\dfrac{1}{15\times16}\right)\)
\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(=\dfrac{3}{8}\)
=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