Đáp án: $x=\pm2$
Giải thích các bước giải:
Ta có :
$\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=\dfrac{x^2-3}{1.2}+\dfrac{x^2-3}{3.4}+...+\dfrac{x^2-3}{99.100}$
$\to \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=(x^2-3)(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100})$
$\to \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=(x^2-3)(\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100})$
$\to \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=(x^2-3)(\dfrac11-\dfrac12+\dfrac13-\dfrac14+..+\dfrac1{99}-\dfrac1{100})$
$\to \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=(x^2-3)((\dfrac11+\dfrac13+\dfrac15+..+\dfrac1{99})-(\dfrac12+\dfrac14+..+\dfrac1{100}))$
$\to \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=(x^2-3)((\dfrac11+\dfrac12+\dfrac13+\dfrac14+\dfrac15+..+\dfrac1{99}+\dfrac1{100})-2(\dfrac12+\dfrac14+..+\dfrac1{100}))$
$\to \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=(x^2-3)((\dfrac11+\dfrac12+\dfrac13+\dfrac14+\dfrac15+..+\dfrac1{99}+\dfrac1{100})-(\dfrac11+\dfrac12+..+\dfrac1{50}))$
$\to \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=(x^2-3)(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100})$
$\to x^2-3=1$
$\to x^2=4$
$\to x=\pm2$
