tìm số TN x biết
(1/1.2.3+1/2.3.4+...+1/8.9.10)x=22/45
đang cần gấp nên mình sẽ cho 2 tick
K
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SN
10 tháng 8 2021
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\dfrac{22}{45}.x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.x=1\)
=> \(x=2\)
Vậy x = 2
Chúc bạn học tốt !!!
LN
1
DD
Đoàn Đức Hà
Giáo viên
25 tháng 1 2021
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(A=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{10-8}{8.9.10}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{11}{45}\).
Phương trình tương đương với:
\(\frac{11}{45}x=\frac{22}{45}\Leftrightarrow x=2\).
D
12 tháng 10 2019
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\cdot\frac{22}{45}=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}=1\)
\(\Rightarrow x=2\)
TT
0
NH
1
TN
16 tháng 3 2016
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\cdot x=\frac{22}{45}\)
\(\Rightarrow1-\frac{1}{10}.x=\frac{22}{45}\)
\(\Rightarrow\frac{9}{10}.x=\frac{22}{45}\)
\(\Rightarrow x=\frac{22}{45}:\frac{9}{10}\)
\(\Rightarrow x=\frac{22}{45}.\frac{10}{9}=\frac{22.10}{45.9}=\frac{44}{81}\)
=>x=\(\frac{44}{81}\)
NT
1
16 tháng 3 2017
\(pt\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}x=1\)
\(\Rightarrow x=2\)
TG
14 tháng 3 2019
\(\left(\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-.........+\frac{1}{8}-\frac{1}{9}-\frac{1}{10}\right)\).x = \(\frac{22}{45}\)
\(\left(1-\frac{1}{10}\right).x=\frac{22}{45}\)
\(\frac{9}{10}.x=\frac{22}{45}\)
\(x=\frac{22}{45}:\frac{9}{10}\)
\(x=\frac{44}{81}\)
(1/1.2.3+1/2.3.4+...+1/8.9.10).x=22/45
<=>(2/1.2.3+2/2.3.4+...+2/8.9.10).x=44/45
<=>(1/1.2-1/2.3+1/2.3-1/3.4+...+1/8.9-1/9.10).x=44/45
<=>(1/1.2-1/9.10).x=44/45
<=>22/45.x=44/45
<=>x=2
vậy x=2
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.............+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+.........+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+............+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{22}{45}\)
\(\Rightarrow x=\frac{22}{45}:\frac{11}{45}\)
\(\Rightarrow x=2\)