Đáp án:
$⇒x=\dfrac{117}{23}$
Giải thích các bước giải:
Em ơi đề sai rồi:$\dfrac{1}{5×6}+..+\dfrac{1}{x×(x+1)}=\dfrac{1}{28}$
$ADCT:\dfrac{k}{n(n+k)}=\dfrac{1}{n}-\dfrac{1}{n+k}$
$⇒\dfrac{1}{5}-\dfrac{1}{6}+..+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{28}$
$⇒\dfrac{1}{5}-\dfrac{1}{x+1}=\dfrac{1}{28}$
$⇒\dfrac{1}{x+1}=\dfrac{1}{5}-\dfrac{1}{28}$
$⇒\dfrac{1}{x+1}=\dfrac{23}{140}$
$⇒x=\dfrac{117}{23}$