x.(4/1.2+4/2.3+4/3.4+…+4/99.100)=33/25

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1. `x . ( 4/( 1 . 2 ) + 4/( 2 . 3 ) + 4/( 3 . 4 ) + … + 4/( 99 . 100 ) ) = 33/25`

   `⇔ x . 4 . ( 1/( 1 . 2 ) + 1/( 2 . 3 ) + 1/( 3 . 4 ) + … + 1/( 99 . 100 ) ) = 33/25`

   `⇔ x . 4 . ( 1/1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + … + 1/99 – 1/100 ) = 33/25`

   `⇔ x . 4 . ( 1 – 1/100 ) = 33/25`

   `⇔       x . 4 . 99/100   = 33/25`

   `⇔                x . 99/25 = 33/25`

   `⇔                        x     = 33/25 : 99/25`

   `⇔                        x     = 1/3`

   Vậy , `x = 1/3 .`

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2. Đáp án:

    `x=1/3`

   Giải thích các bước giải:

   `x.[4/1.2+4/2.3+4/3.4+…+4/99.100]=33/25`

   `<=>x.[4.(1-1/2+1/2-1/3+1/3-1/4+…+1/99-1/100)]=33/25`

   `<=>x.[4.(1-1/100)]=33/25`

   `<=>x.[4 . 99/100]=33/25`

   `<=>x. 99/25=33/25`

   `<=>x=33/25÷99/25`

   `<=>x=33/25xx25/99`

   `<=>x=1/3`

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