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Question

CMR:

$\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}$

Nguyễn Thanh Hằng · Accepted Answer

Ta có :

$\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.........+\dfrac{1}{49.50}$

$=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{49}-\dfrac{1}{50}$

$=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+......+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{50}\right)$

$=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{50}\right)$

$=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+.....+\dfrac{1}{25}\right)$

$=\dfrac{1}{26}+\dfrac{1}{27}+......+\dfrac{1}{50}$

Vậy ...Câu trả lời: Ta có :

$\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.........+\dfrac{1}{49.50}$

$=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{49}-\dfrac{1}{50}$

$=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+......+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{50}\right)$

$=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{50}\right)$

$=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+.....+\dfrac{1}{25}\right)$

$=\dfrac{1}{26}+\dfrac{1}{27}+......+\dfrac{1}{50}$

Vậy ...

Mashiro Shiina · Answer

Đặt:

$PHUCDZ=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}$

$PHUCDZ=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}$

$PHUCDZ=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)$

$PHUCDZ=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)$

$PHUCDZ=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)$

$PHUCDZ=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}$

Đặt $PHUCMAXDZ=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}$

$PHUCDZ=PHUCMAXDZ$ vậy ta có $đpcm$Câu trả lời: Đặt:

$PHUCDZ=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}$

$PHUCDZ=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}$

$PHUCDZ=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)$

$PHUCDZ=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)$

$PHUCDZ=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)$

$PHUCDZ=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}$

Đặt $PHUCMAXDZ=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}$

$PHUCDZ=PHUCMAXDZ$ vậy ta có $đpcm$

Heartilia Hương Trần · Answer

Ta có : $\dfrac{1}{1.2}$+$\dfrac{1}{3.4}$ + $\dfrac{1}{5.6}$ +...+ $\dfrac{1}{49.50}$

= (1-$\dfrac{1}{2}$) + ($\dfrac{1}{3}$-$\dfrac{1}{4}$) + ($\dfrac{1}{5}$-$\dfrac{1}{6}$) + ... + ($\dfrac{1}{49}$-$\dfrac{1}{50}$)

= (1+$\dfrac{1}{3}$ +$\dfrac{1}{5}$+....+$\dfrac{1}{49}$) - ( $\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{6}$+...+$\dfrac{1}{50}$)

=(1+$\dfrac{1}{3}$+$\dfrac{1}{5}$+...+$\dfrac{1}{49}$) - 2($\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{6}$+...+$\dfrac{1}{50}$)

= (1+$\dfrac{1}{2}$+$\dfrac{1}{3}$+...+$\dfrac{1}{50}$) - (1+$\dfrac{1}{2}$+$\dfrac{1}{3}$+...+$\dfrac{1}{25}$)

=$\dfrac{1}{26}$+$\dfrac{1}{27}$+...+$\dfrac{1}{50}$ (đpcm)Câu trả lời: Ta có : $\dfrac{1}{1.2}$+$\dfrac{1}{3.4}$ + $\dfrac{1}{5.6}$ +...+ $\dfrac{1}{49.50}$

= (1-$\dfrac{1}{2}$) + ($\dfrac{1}{3}$-$\dfrac{1}{4}$) + ($\dfrac{1}{5}$-$\dfrac{1}{6}$) + ... + ($\dfrac{1}{49}$-$\dfrac{1}{50}$)

= (1+$\dfrac{1}{3}$ +$\dfrac{1}{5}$+....+$\dfrac{1}{49}$) - ( $\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{6}$+...+$\dfrac{1}{50}$)

=(1+$\dfrac{1}{3}$+$\dfrac{1}{5}$+...+$\dfrac{1}{49}$) - 2($\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{6}$+...+$\dfrac{1}{50}$)

= (1+$\dfrac{1}{2}$+$\dfrac{1}{3}$+...+$\dfrac{1}{50}$) - (1+$\dfrac{1}{2}$+$\dfrac{1}{3}$+...+$\dfrac{1}{25}$)

=$\dfrac{1}{26}$+$\dfrac{1}{27}$+...+$\dfrac{1}{50}$ (đpcm)

Violympic toán 7

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