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
$\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\right)x=\dfrac{2012}{51}+\dfrac{2012}{52}+..+\dfrac{2012}{100}$
$\to \left(\dfrac1{51}+\dfrac{1}{52}+...+\dfrac1{100}\right)x=2012\left(\dfrac1{51}+\dfrac{1}{52}+...+\dfrac1{100}\right)$
$\to x=2012$
