(1/1.2+1/3.4+1/4.5+1/5.6+⋯+1/99.100).x =2012/51+2012/52+2012/53+⋯+2012/100 Các bạn ơi giúp mk làm bài với ạ Mk đang rất cần Thanks ạ

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1. Đáp án: $x=2012$

   Giải thích các bước giải:

   Sửa đề:

   $\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+…+\dfrac{1}{99.100}\right)x=\dfrac{2012}{51}+\dfrac{2012}{52}+..+\dfrac{2012}{100}$

   Ta có:
   $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+…+\dfrac{1}{99.100}$

   $\to A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+\dfrac{6-5}{5.6}+…+\dfrac{100-99}{99.100}$

   $\to A=\dfrac11-\dfrac12+\dfrac13-\dfrac14+\dfrac15-\dfrac16+…+\dfrac1{99}-\dfrac1{100}$

   $\to A=\left(1+\dfrac13+\dfrac15+…+\dfrac1{99}\right)-\left(\dfrac12+\dfrac14+\dfrac16+…+\dfrac1{100}\right)$

   $\to A=\left(1+\dfrac13+\dfrac15+…+\dfrac1{99}\right)+\left(\dfrac12+\dfrac14+\dfrac16+…+\dfrac1{100}\right)-2\left(\dfrac12+\dfrac14+\dfrac16+…+\dfrac1{100}\right)$

   $\to A=\left(1+\dfrac12+\dfrac13+\dfrac14+…+\dfrac1{99}+\dfrac1{100}\right)-\left(1+\dfrac12+\dfrac13+…+\dfrac1{50}\right)$

   $\to A=\dfrac1{51}+\dfrac{1}{52}+…+\dfrac1{100}$

   Mà 

   $\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+…+\dfrac{1}{99.100}\right)x=\dfrac{2012}{51}+\dfrac{2012}{52}+..+\dfrac{2012}{100}$

   $\to \left(\dfrac1{51}+\dfrac{1}{52}+…+\dfrac1{100}\right)x=2012\left(\dfrac1{51}+\dfrac{1}{52}+…+\dfrac1{100}\right)$

   $\to x=2012$

   Trả lời