Đáp án:
Đặt $A=\dfrac{1}{1×2}+\dfrac{1}{3×4}+\dfrac{1}{5×6}+...+\dfrac{1}{49×50}$
Áp dụng tính chất:$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$
$A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{49}-\dfrac{1}{50}$
$A=(1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{49}+\dfrac{1}{50})-2(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{25})$
$A=1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{49}+\dfrac{1}{50}-(1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25})$
$A=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+....+\dfrac{1}{50}$
