[Toán 7]Tính:

S

su10112000a

A=$\frac{1}{99.97}$-$\frac{1}{97.95}$-$\frac{1}{95.93}$-...-$\frac{1}{5.3}$-$\frac{1}{3.1}$
A=-($\frac{1}{1.3}$+$\frac{1}{5.3}$+...+$\frac{1}{95.93}$+$\frac{1}{97.95}$)+
$\frac{1}{99.97}$
A=$\frac{-1}{2}$.($\frac{1}{1}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{5}$+...+$\frac{1}{95}$-$\frac{1}{97}$)

+$\frac{1}{99.97}$
A=
$\frac{-1}{2}$.($\frac{1}{1}$-$\frac{1}{97}$)+$\frac{1}{99.97}$
A=$\frac{-48}{97}$
+$\frac{1}{99.97}$
A=$\frac{-4751}{9603}$
không biết có đúng không
:D:D:D:D:D
 
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D

demon311

$-2A=\dfrac{ 2}{1.3}+\dfrac{ 2}{3.5}+...+\dfrac{ 2}{95.97}-\dfrac{ 2}{97.99} \\
=1-\dfrac{ 1}{3}+\dfrac{ 1}{3}-\dfrac{ 1}{5}+...+\dfrac{ 1}{95}-\dfrac{ 1}{97}-\dfrac{ 1}{97}+\dfrac{ 1}{99} = 1+\dfrac{ 1}{99}-\dfrac{ 2}{97}$
Chắc su làm đúng rồi
 
D

duc_2605

$A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}$
$A=-(\dfrac{1}{97.99}+\dfrac{1}{95.97}+\dfrac{1}{93.95}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3})$
$-1/2.A=\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{1}-\dfrac{1}{3}$
-1/2A = 1 - 1/99 + 2/97= -9502/9603
=> A

 
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