Tìm x biết :
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{1}{X\left(X+2\right)}=\frac{2}{9}\)
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
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DP
24 tháng 6 2017
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{19.21}\right).x=\frac{9}{7}\)
\(\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{21-19}{19.21}\right).x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\frac{2}{7}\right].x=\frac{9}{7}\)
\(\frac{1}{7}.x=\frac{9}{7}\)
\(\Rightarrow x=\frac{9}{7}\div\frac{1}{7}=9\)
24 tháng 6 2017
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\frac{2}{7}x=\frac{9}{7}\)
\(\Leftrightarrow x=\frac{9}{2}\)
29 tháng 1 2016
Bước 1: \(\left(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{19\times21}\right)=\frac{1}{7}\)
Bước 2: \(x=\frac{9}{7}\div\frac{1}{7}=9\)
3 tháng 5 2018
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
PV
24 tháng 4 2016
* ĐK: \(x\ne0\)
Đề ra ...<=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{1}{9}\)
<=> \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{6}-\frac{1}{x+1}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{x+1}\left(1-\frac{1}{x}\right)=\frac{1}{6}-\frac{1}{9}\)
<=> \(\frac{x-1}{x\left(x+1\right)}=\frac{1}{36}\)
<=> \(\frac{x-1}{x\left(x-1\right)}=\frac{x-1}{36.\left(x-1\right)}\)
=> x(x-1) = 36. (x-1) => x =36
TN
24 tháng 4 2016
\(\frac{2}{2}.\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x+\left(x+1\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)
\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
x+1=18
x=18-1
x=17
25 tháng 3 2019
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+....+\frac{2}{x}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{6}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{3}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{3}{9}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
25 tháng 3 2019
câu a khó quá.Để nghĩ.
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{21\cdot2}+\frac{2}{28\cdot2}+\frac{2}{36\cdot2}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x-1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)
\(\Rightarrow9\left(x-5\right)=6x+6\)
\(\Rightarrow9x-45=6x+6\)
\(\Rightarrow9x-6x=51\)
\(\Rightarrow3x=51\)
Tới đây bí:v
a)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+2\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+2}=\frac{2}{9}:2\)
\(\frac{1}{x+2}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+2}=\frac{1}{18}\)
=>x+2=18
=>x=16
b tương tự nhân nó với 1/2
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