P=1/99 – 1/99.98 – 1/98.97 – 1/97.96 – ……. – 1/3.2 – 1/2.1 October 30, 2020 by Ben Gia P=1/99 – 1/99.98 – 1/98.97 – 1/97.96 – ……. – 1/3.2 – 1/2.1
$P=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-…-\dfrac{1}{3.2}-\dfrac{1}{2.1}$ $⇔P=-\left(-\dfrac{1}{99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+\dfrac{1}{97.96}+…+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)$ $⇔P=-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+…+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}-\dfrac{1}{99}\right)$ $⇔P=-\left(\dfrac{1}{1}-\dfrac{1}{99}-\dfrac{1}{99}\right)$ `⇔P=-97/99` Học tốt ! Reply
`P=1/99 – 1/(99.98) – 1/(98.97) – 1/(97.96) – … – 1/(3.2) – 1/(2.1)` `P=1/99-1/98+1/99-1/87+1/98-…-1+1/2` `P=2/99-1` `P=-97/99` Reply
$P=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-…-\dfrac{1}{3.2}-\dfrac{1}{2.1}$
$⇔P=-\left(-\dfrac{1}{99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+\dfrac{1}{97.96}+…+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)$
$⇔P=-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+…+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}-\dfrac{1}{99}\right)$
$⇔P=-\left(\dfrac{1}{1}-\dfrac{1}{99}-\dfrac{1}{99}\right)$
`⇔P=-97/99`
Học tốt !
`P=1/99 – 1/(99.98) – 1/(98.97) – 1/(97.96) – … – 1/(3.2) – 1/(2.1)`
`P=1/99-1/98+1/99-1/87+1/98-…-1+1/2`
`P=2/99-1`
`P=-97/99`