Toán Violympic.

A

anconan5a

$\frac{1}{3}$+$\frac{1}{5}$+...+$\frac{1}{99}$+$\frac{1}{143}$
=$\frac{1}{1.3}$+$\frac{1}{3.5}$+...+$\frac{1}{9.11}$+$\frac{1}{11.13}$
=1-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{5}$+...+$\frac{1}{9}$-$\frac{1}{11}$+$\frac{1}{13}$
=1-$\frac{1}{13}$
=$\frac{12}{13}$
p/s:Máy mình bị lỗi, bạn thông cảm
 
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T

tayhd20022001


Tính:
$\dfrac{1}{3}$+$\dfrac{1}{15}$+..........+$\dfrac{1}{99}$+$\dfrac{1}{143}$
Giải
Ta có :
$\dfrac{1}{3}$+$\dfrac{1}{15}$+..........+$\dfrac{1}{99}$+$\dfrac{1}{143}$
\Rightarrow $\dfrac{1}{1.3}$+$\dfrac{1}{3.5}$+..........+ $\dfrac{1}{9.11}$+$\dfrac{1}{11.13}$
\Rightarrow 1 - $\dfrac{1}{3}$ + $\dfrac{1}{3}$ - $\dfrac{1}{5}$+...+ $\dfrac{1}{9}$- $\dfrac{1}{11}$+ $\dfrac{1}{11}$- $\dfrac{1}{13}$
\Rightarrow = 1- $\dfrac{1}{13}$
Vậy \Rightarrow 1- $\dfrac{1}{13}$ =$\dfrac{12}{13}$
 
V

vuivemoingay

[TEX]\frac{1}{3}+\frac{1}{15}+..........+\frac{1}{99}+\frac{1}{143}[/TEX]

[TEX]= \frac{1}{2}.\frac{2}{1.3} + \frac{1}{2}.\frac{2}{3.5} + ... + \frac{1}{2}.\frac{2}{9.11} + \frac{1}{2}.\frac{2}{11.13}[/TEX]

[TEX]= \frac{1}{2} . (\frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{9.11} + \frac{2}{11.13})[/TEX]

[TEX]= \frac{1}{2}. (1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + ... + \frac{1}{9} - \frac{1}{11} + \frac{1}{11} - \frac{1}{13})[/TEX]

[TEX]= \frac{1}{2}. (1 - \frac{1}{13})[/TEX]

[TEX]= \frac{1}{2}.\frac{12}{13}[/TEX]

[TEX]= \frac{6}{13}[/TEX]
 
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