Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=1999/2000
B=199/200
C=511/512
hok tốt
Đáp án
mình lười trình bày cách làm lém, để đáp án thui nha
A = \(\frac{1999}{2000}\)
B = \(\frac{199}{200}\)
C = \(\frac{511}{512}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{2012\times2014}\)
\(=\frac{1}{2}\times(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2012\times2014})\)
\(=\frac{1}{2}\times(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2012}-\frac{1}{2014})\)
\(=\frac{1}{2}\times(\frac{1}{2}-\frac{1}{2014})\)
\(=\frac{1}{2}\times(\frac{1007}{2014}-\frac{1}{2014})\)
\(=\frac{1}{2}\times\frac{503}{1007}\)
\(=\frac{503}{2014}\)
Ta có ; \(\frac{1}{2}=\frac{1007}{2014}\)
Vậy A bé hơn B
Chúc bạn học tốt
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{1}-\frac{1}{8}\)
\(=\frac{7}{8}\)'
\(\frac{5}{1.6}\)+ \(\frac{5}{6.11}\)+ .........+\(\frac{5}{501.506}\)
=\(\frac{1}{1.6}+\frac{1}{6.11}+.....+\frac{1}{501.506}\)
=\(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+......+\frac{1}{501}-\frac{1}{506}\)
=\(\frac{1}{1}-\frac{1}{506}\)
= tự tính nha
= 1/2+1/4+....+1/512+1/512 - 1/512
= 1/2+1/4+....+1/256+1/256 - 1/512
........
= 1/2+1/2 - 1/512 = 1-1/512 = 511/512
k mk nha
Gọi biểu thức trên là A
Ta có :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}+\frac{1}{256}-\frac{1}{256}\)
\(2A=1+A-\frac{1}{256}\)
\(2A=A+1-\frac{1}{256}\)
\(2A-A=\frac{255}{256}\)
\(A=\frac{255}{256}\)
Gọi \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right]-\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right]\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^8}\)
\(A=1-\frac{1}{2^8}=1-\frac{1}{256}=\frac{255}{256}\)
Ta có: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
= \(\frac{1}{2}+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+....+\left(\frac{1}{64}-\frac{1}{128}\right)\)
=\(\frac{1}{2}+\frac{1}{2}-\frac{1}{128}\)
\(=1-\frac{1}{128}=\frac{127}{128}\)
MẤY CÂU KHÁC THÌ SAO?