\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=1-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=1-\dfrac{1}{50}=\dfrac{49}{50}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}.\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{49}-\dfrac{1}{50}.\)
\(=\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+.....+\left(-\dfrac{1}{49}+\dfrac{1}{49}\right)+\left(1-\dfrac{1}{50}\right).\)\(=0+0+0+.....+0+\left(1-\dfrac{1}{50}\right).\)
\(=1-\dfrac{1}{50}.\)
=\(\dfrac{49}{50}.\)
CHÚC BN HỌC TỐT!!! :) :) :)
Đừng quên bình luận nếu bài tớ sai nhé!!!
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=1-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
