Câu hỏi của Nguyễn Ngọc Gia Hân

Question

Thu gọn :

A = $\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+.......+\dfrac{9998}{9999}$

Mới vô · Accepted Answer

$A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+...+\dfrac{9998}{9999}\
=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{9999}\right)\
=\left(1+1+1+...+1\right)\left(\text{có 50 số 1}\right)-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\
=50\cdot1-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\right)\
=50-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\
=50-\left(1-\dfrac{1}{101}\right)\
=50-1+\dfrac{1}{101}\
=49+\dfrac{1}{101}\
=\dfrac{4949+1}{101}\
=\dfrac{4950}{101}$Câu trả lời: $A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+...+\dfrac{9998}{9999}\
=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{9999}\right)\
=\left(1+1+1+...+1\right)\left(\text{có 50 số 1}\right)-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\
=50\cdot1-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\right)\
=50-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\
=50-\left(1-\dfrac{1}{101}\right)\
=50-1+\dfrac{1}{101}\
=49+\dfrac{1}{101}\
=\dfrac{4949+1}{101}\
=\dfrac{4950}{101}$

Đại số lớp 6

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)