Tìm `y` biết: `1/4+1/28+1/70+1/130+…+1/(y(y+3))=34/103` `1/3+1/6+1/10+1/15+…+1/((y(y+1))/2)=2009/2011`

0 bình luận về “Tìm `y` biết: `1/4+1/28+1/70+1/130+…+1/(y(y+3))=34/103` `1/3+1/6+1/10+1/15+…+1/((y(y+1))/2)=2009/2011`”

1. a)1/4+1/28+1/70+1/130+…+1/y(y+3)=34/103

   ⇔1/1*4+1/4*7+1/7*10+1/10*13+….+1/y*(y+3)=34/103

   ⇔3(1/1*4+1/4*7+1/7*10+1/10*13+….+1/y*(y+3)=102/103

   ⇔3/1*4+3/4*7+3/7*10+….+3/y(y+3)=102/103

   ⇔1-1/4+1/4-1/7+1/7+1/10+…+1/y-1/y+3=102/103

   ⇔1-1/y+3=102/103

   ⇔1/y+3=1-102/103=1/103

   ⇔y+3=103

   ⇔y=100

   b)1/3+1/6+1/10+1/15+…+1/y(y+1)/2=2009/2011

   ⇔1/2(1/3+1/6+1/10+1/15+…+1/y(y+1)/2)=2009/2011*1/2=2009/4022

   ⇔1/2*3+1/3*4+…+1/y(y+1)=2009/4022

   ⇔1/2-1/3+1/3+1/4+….+1/y-1/y+1=2009/4022

   ⇔1/2-1/y+1=2009/4022

   ⇔1/y+1=1/2-2009/4022=2/4022=1/2011

   ⇔y+1=2011

   ⇔y=2010

    

   Bình luận

2. Đáp án:

   $a/$ `1/4 + 1/28 + 1/70 + 1/130 + .. + 1/(y (y + 3) ) = 34/103`

   `⇔ 1/(1 . 4) + 1/(4 . 7) + 1/(7 . 10) + 1/(10 . 14) + … + 1/(y (y + 3)) = 34/103`

   `⇔ 1/3 [1 – 1/4 + 1/4 – 1/7 + 1/7 – 1/10 + 1/10 – 1/14 + … + 1/y – 1/(y + 3)] = 34/103`

   `⇔ 1/3  [1  + ( – 1/4 + 1/4 – 1/7 + 1/7 – 1/10 + 1/10 – 1/14 + … + 1/y) – 1/(y +3)] = 34/103`

   `⇔ 1/3 [1 –   1/(y + 3)] = 34/103`

   `⇔ 1-  1/(y + 3) = 102/103`

   `⇔ 1/(y + 3) = 1/103`

   `⇔ 103 = y + 3`

   `⇔ y = 100`

   `text{Vậy y =100}`

   $b/$ `1/3 + 1/6 + 1/10 + 1/15 + .. + 1/( (y (y + 1) )/2) = 2009/2011`

   `⇔ 1/(2 . 3) + 1/(3 . 4) + 1/(4 . 5) + … + 1/(y (y + 1) ) = 2009/4022`

   `⇔ 1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + .. + 1/y – 1/(y + 1) = 2009/4022`

   `⇔ 1/2 + (- 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + .. + 1/y) – 1/(y + 1) = 2009/4022`

   `⇔ 1/2 – 1/(y + 1) = 2009/4022`

   `⇔ 1/(y + 1) = 1/2011`

   `⇔ y + 1 = 2011`

   `⇔ y = 2010`

   `text{Vậy y = 2010}`

    

   Bình luận