Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(A=\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{210}\)
=> \(\dfrac{1}{2}A=\dfrac{1}{2}\left(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{210}\right)\text{}\)
\(=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{420}\)
\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{20.21}\)
\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{20}-\dfrac{1}{21}\)
\(=\dfrac{1}{6}-\dfrac{1}{21}\)
\(=\dfrac{5}{42}\)
Vậy \(A=\dfrac{5}{42}\)
=\(\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}+\dfrac{1}{9.5}\)=\(\dfrac{1}{3}+\dfrac{1}{5}\)
Gọi A = \(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{2}.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{3}-\dfrac{1}{10}\)
\(\dfrac{1}{2}\)A = \(\dfrac{7}{30}\)
A = \(\dfrac{7}{30}:\dfrac{1}{2}\)
A = \(\dfrac{7}{15}\)
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{2}{9}\)
<=> \(\dfrac{1}{6.7:2}+\dfrac{1}{7.8:2}+\dfrac{1}{8.9:2}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{2}{9}\)
<=> \(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
<=> \(2\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)
<=> \(2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
<=> \(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
<=> \(\dfrac{1}{x+1}=\dfrac{1}{18}\)
<=> x + 1 = 18
<=> x = 17
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{x.\left(x+1\right):2}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{2}{9}.2=\dfrac{4}{9}\)\(\Leftrightarrow\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{4}{9}\)\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{4}{9}\)\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{4}{9}-\dfrac{1}{6}=\dfrac{5}{8}\)
\(\Leftrightarrow\left(1.8\right)=5\left(x+1\right)\)
\(\Leftrightarrow8=5x+5\)
\(\Leftrightarrow5x=8-3=5\)
\(\Leftrightarrow x=5:5\)
\(\Leftrightarrow x=1\)
A =\(2.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{156}\right)\)
A =\(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+..........+\dfrac{1}{12.13}\right)\)
A =2.\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
A=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\cdot\left[\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{x\left(x+1\right)}\right]=\dfrac{2}{9}\\ \dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2}{9}:2\\ \dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\\ \dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{4}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{18}\\ x+1=18\\ x=17\)
Vậy x = 17
a, sai đề
b, \(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Rightarrow\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\) ( nhân cả 2 vế với \(\dfrac{1}{2}\) )
\(\Rightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
Câu a thiếu đề rồi bạn ơi mik giải câu b đây:
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{x+2}\right)=\dfrac{2}{9}\)
\(\dfrac{1}{6}-\dfrac{1}{x+2}=\dfrac{2}{9}:2\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
\(\dfrac{1}{2}N=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)
N=\(\dfrac{2}{5}:\dfrac{1}{2}=\dfrac{4}{5}\)
M = 1/21 + 1/28+1/36+...+1/465
= 2/42+2/56+2/72+...+2/930
= 2.( 1/6.7 + 1/7.8 + 1/ 7.9 + ... + 1/30.31)
= 2.( 1/6-1/7+1/7-1/8+...+1/30-1/31)
= 2.(1/6 - 1/31) = 2.25/186 = 25/92