[Toán 7]

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shinata

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pinkylun

$E=\dfrac{1}{1+2} + \dfrac{1}{1+2+3} + \dfrac{1}{1+2+3+4} +...\dfrac{1}{1+2+3..+24}$

$E=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{24.25}$

$E=2(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{24}-\dfrac{1}{25}$

$E=2.(\dfrac{1}{2}-\dfrac{1}{25})$

$E=2.\dfrac{23}{50}=\dfrac{23}{25}$

$D=\dfrac{10}{56} + \dfrac{10}{140} + \dfrac{10}{260} + ....+ \dfrac{10}{1400}$

$D=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}$

$ D=\dfrac{5}{3}(\dfrac{3}{4.7}+ \dfrac{3}{7.10}+ \dfrac{3}{10.13}+...+\dfrac{3}{25.28})$

$D=\dfrac{5}{3}(\dfrac{1}{4}-\dfrac{1}{28})$

$D=\dfrac{5}{3}(\dfrac{3}{14})$

$D=\dfrac{5}{14}$

$=>\dfrac{D}{E}=...$


 
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N

ngocsangnam12

pinkylun said:
$E=\dfrac{1}{1+2} + \dfrac{1}{1+2+3} + \dfrac{1}{1+2+3+4} +...\dfrac{1}{1+2+3..+24}$

$E=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{24.25}$

$E=2(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{24}-\dfrac{1}{25}$

$E=2.(\dfrac{1}{2}-\dfrac{1}{25})$

$E=2.\dfrac{23}{50}=\dfrac{23}{25}$

$D=\dfrac{10}{56} + \dfrac{10}{140} + \dfrac{10}{260} + ....+ \dfrac{10}{1400}$


p.s: gửi trc theo lời anh hay em j đó cua bài này
Bấm để xem đầy đủ nội dung ...

Còn D thì thu gọn mỗi vế đi 2 lần *Vào muộn rồi*

$D=\dfrac{10}{56} + \dfrac{10}{140} + \dfrac{10}{260} + ....+ \dfrac{10}{1400}$
$D= \frac{5}{28}+\frac{5}{70}+\frac{5}{130}+....+\frac{5}{700}$
$D=\frac{5}{4.7}+\frac{5}{7.10}+.......+\frac{5}{25.28}$
$\frac{3}{5}D=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{25}-\frac{1}{28}$
$\frac{3}{5}D=\frac{1}{4}-\frac{1}{28}=\frac{3}{14}$
$D=\frac{3}{14}:\frac{3}{5}=\frac{5}{14}$
Rôi lấy chia cho nhau đó
 
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