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23 tháng 7 2017

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)

\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+......+\frac{2}{8.9.10}\right).x=\frac{22}{45}\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{22}{45}\)

\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{22}{45}\)

\(\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)

\(\frac{11}{45}.x=\frac{22}{45}\)

\(x=\frac{22}{45}:\frac{11}{45}\)

\(x=2\)

31 tháng 1 2016

Đặt B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{8.9.10}\)

=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{8.9.10}\)

=> 2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\)

=> 2B = \(\frac{1}{1.2}-\frac{1}{9.10}\)

2B = \(\frac{22}{45}\)

B = \(\frac{22}{45}:2\)

=> B = \(\frac{11}{45}\)

Ta có : \(\frac{11}{45}.x=\frac{22}{45}\)

=> x = \(\frac{22}{45}:\frac{11}{45}\)

=> x = \(\frac{2}{1}\)

5 tháng 12 2015

\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)

\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)

\(\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\)

\(\frac{22}{45}.x=\frac{44}{45}\)

x =2

 

5 tháng 12 2015

X = 2 

9 tháng 10 2020

e. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Rightarrow x=15\)

f. \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{22}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)

\(\Rightarrow x=2\)

21 tháng 2 2016

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)

\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\Leftrightarrow\frac{22}{45}.x=\frac{44}{45}\Leftrightarrow x=2\)

Vậy x=2

1 tháng 1 2017

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{44}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{44}{45}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{44}{45}\)

\(\Rightarrow\frac{22}{45}.x=\frac{44}{45}\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

1 tháng 1 2017

2

AH
Akai Haruma
Giáo viên
12 tháng 4 2022

Lời giải:

Đặt $A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}$

$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{8.9.10}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{10-8}{8.9.10}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}$

$=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}$

$A=\frac{11}{45}$

$Ax=\frac{11}{45}x=\frac{22}{45}$

$x=\frac{22}{45}: \frac{11}{45}=2$