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(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
= 1/ 2 ( 1/1.2 - 1/2.3 +1/2.3-1/3.4+...+1/8.9-1/9.10 ) x = 22/45
= 1/2 .(1/2-1/90)x = 22/45
= 1/2 . 22/45. x=22/45
x = 22/45:( 22/45 .1/2)
x =2
duyệt
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{23}{45}\)
=> \(\left[\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\right)\right]x=\frac{23}{45}\)
=>\(\left[\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)
=> \(\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)
=> \(\left[\frac{1}{2}\cdot\frac{22}{45}\right]x=\frac{23}{45}\)
=> \(\frac{11}{45}x=\frac{23}{45}\)
=> \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{45}\cdot\frac{45}{11}=\frac{23}{11}\)
Vậy x = 23/11
Ez :))
(1/1.2.3 + 1/2.3.4 + ... + 1/8.9.10).x = 23/45
(2/1.2.3 + 2/2.3.4 + ... + 2/8.9.10).x = 2.23/45
(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/8.9 - 1/9.10).x = 46/45
(1/1.2 - 1/9.10).x = 46/45
(1/2 - 1/90).x = 46/45
(45/90 - 1/90).x = 46/45
22/45.x = 46/45
x = 46/45 : 22/45
x = 46/45 . 45/22
x = 23/11
=2.(1/1-1/2-1/3+1/2-1/3-1/4+...+1/8-1/9-1/10).x=2.23/45
=(1/1-1/10).x=46/45
=9/10.x=46/45
=x=46/45:9/10
=x=92/81:2=46/81.
k cho mình nha!!!"
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}.\frac{44}{90}.x=\frac{23}{45}\Rightarrow\frac{11}{45}.x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)