Câu hỏi của Hồ Hà Lan

Question

em xin giải chi tiết bài này em cảm ơn ạaaTT

Hà Quang Minh · Accepted Answer

$a,A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\
=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\
=1-\dfrac{1}{100}\
=\dfrac{99}{100}\
b,B=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{97\times100}\
=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\
=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{100}\right)\
=\dfrac{1}{3}\cdot\dfrac{99}{100}\
=\dfrac{33}{100}$$c,C=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\
=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^7}\
\Rightarrow2C=1+\dfrac{1}{2}+...+\dfrac{1}{2^6}\
\Rightarrow2C-C=1-\dfrac{1}{2^7}\
\Rightarrow C=\dfrac{127}{128}$Câu trả lời: $a,A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\
=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\
=1-\dfrac{1}{100}\
=\dfrac{99}{100}\
b,B=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{97\times100}\
=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\
=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{100}\right)\
=\dfrac{1}{3}\cdot\dfrac{99}{100}\
=\dfrac{33}{100}$$c,C=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\
=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^7}\
\Rightarrow2C=1+\dfrac{1}{2}+...+\dfrac{1}{2^6}\
\Rightarrow2C-C=1-\dfrac{1}{2^7}\
\Rightarrow C=\dfrac{127}{128}$

Hà Quang Minh · Answer

$d,D=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{99\times100}\
=2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\
=2\cdot\left(1-\dfrac{1}{100}\right)\
=\dfrac{99}{55}$$e,E=\dfrac{1}{10}+\dfrac{1}{40}+...+\dfrac{1}{340}\
=\dfrac{1}{2\times5}+\dfrac{1}{5\times8}+...+\dfrac{1}{17\times20}\
=\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\
=\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\
=\dfrac{1}{3}\cdot\dfrac{9}{20}\
=\dfrac{3}{20}$Câu trả lời: $d,D=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{99\times100}\
=2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\
=2\cdot\left(1-\dfrac{1}{100}\right)\
=\dfrac{99}{55}$$e,E=\dfrac{1}{10}+\dfrac{1}{40}+...+\dfrac{1}{340}\
=\dfrac{1}{2\times5}+\dfrac{1}{5\times8}+...+\dfrac{1}{17\times20}\
=\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\
=\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\
=\dfrac{1}{3}\cdot\dfrac{9}{20}\
=\dfrac{3}{20}$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP