chứng minh rằng: 1/1.2+1/3.4+1/5.6+……+1/49.50=1/26+1/27+1/28+……+1/50

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1. Giải thích các bước giải:

   `1/(1.2)`+`1/(3.4)`+`1/(5.6)`+……+`1/(49.50)`=`1/26`+`1/27`+`1/28`+……+`1/50`

   =`1/1`-`1/2`+`1/3`-`1/4`+…………….+`1/49`-`1/50`

   =(`1/1`+`1/3`+……..+`1/49`)-(`1/2`+`1/4`+……..+`1/50`)

   =(`1/1`+`1/2`+`1/3`+……….+`1/49`+`1/50`) – 2(`1/2`+`1/4`+……..+`1/50`)

   =`1/1`+`1/2`+`1/3`+……….+`1/49`+`1/50` – `1/1` – `1/2` – ……….. – `1/25`

   =`1/26`+`1/27`+………….+`1/50`

   ⇒ `1/(1.2)`+`1/(3.4)`+`1/(5.6)`+……+`1/(49.50)`=`1/26`+`1/27`+`1/28`+……+`1/50`

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2. `Đáp` `án` + `Giải` `thích“các` `bước` `giải`:

   `$\frac{1}{1.2}$`+`$\frac{1}{3.4}$`+`$\frac{1}{5.6}$`+…+`$\frac{1}{49.50}$`=`$\frac{1}{26}$`+`$\frac{1}{27}$`+`$\frac{1}{28}$`+…+`$\frac{1}{50}$`

   =`1`- `$\frac{1}{2}$`+`$\frac{1}{3}$`-`$\frac{1}{4}$`+`$\frac{1}{5}$`-`$\frac{1}{6}$`+…+`$\frac{1}{49}$`-`$\frac{1}{50}$`

   =(`1`+`$\frac{1}{3}$`+`$\frac{1}{5}$`+…+`$\frac{1}{49}$`) – (`$\frac{1}{2}$`+`$\frac{1}{4}$`+`$\frac{1}{6}$`+…`$\frac{1}{50}$`)

   =(`1`+`$\frac{1}{3}$`+`$\frac{1}{5}$`+…+`$\frac{1}{50}$`) – `2`(`$\frac{1}{2}$`+`$\frac{1}{4}$`+`$\frac{1}{6}$`+…`$\frac{1}{50}$`)

   =(`1`+`$\frac{1}{3}$`+`$\frac{1}{5}$`+…+`$\frac{1}{50}$`) – (`$\frac{1}{2}$`+`$\frac{1}{3}$`+`$\frac{1}{5}$`+…`$\frac{1}{25}$`)

   =`$\frac{1}{26}$`+`$\frac{1}{27}$`+`$\frac{1}{28}$`+…+`$\frac{1}{50}$`

                                                 `XIN` `HAY` `NHẤT` 

    

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