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\(B=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{18\cdot20}\)
\(B=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\)
\(B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\)
\(B=\frac{1}{2}-\frac{1}{20}\)
\(B=\frac{9}{20}\)
=))
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
Lời giải :
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
ko chép lại đề :
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ......... + \(\frac{1}{98}\)- \(\frac{1}{99}\)+ \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
= 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
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{1}-\frac{1}{8}\)
\(=\frac{7}{8}\)'
\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{2012\times2014}\)
\(=\frac{1}{2}\times(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2012\times2014})\)
\(=\frac{1}{2}\times(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2012}-\frac{1}{2014})\)
\(=\frac{1}{2}\times(\frac{1}{2}-\frac{1}{2014})\)
\(=\frac{1}{2}\times(\frac{1007}{2014}-\frac{1}{2014})\)
\(=\frac{1}{2}\times\frac{503}{1007}\)
\(=\frac{503}{2014}\)
Ta có ; \(\frac{1}{2}=\frac{1007}{2014}\)
Vậy A bé hơn B
Chúc bạn học tốt
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 \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right]-\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right]\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^8}\)
\(A=1-\frac{1}{2^8}=1-\frac{1}{256}=\frac{255}{256}\)
Xin lỗi mk nhầm
đề là:
\(1\cdot2\cdot3\cdot4\cdot...\cdot99999999999+\left(\frac{1}{2}+\frac{2}{1}+0,5-1+3-5\right)\)
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 = \(\frac{511}{512}\)
a = 1999/2000
b = 199/200
c = 511/512
A=1999/2000
B=199/200
C=511/512
#HT#
a) 1999/2000
b)199/200
c)511/512
A. \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{2x4}+..+\frac{1}{1999x2000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(=1-\frac{1}{2000}\)
\(=\frac{1999}{2000}\)
B. \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..+\frac{1}{39800}\)
\(=\frac{1}{1x2}+\frac{1}{2x3}+..+\frac{1}{190x200}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1-\frac{1}{200}\)
\(=\frac{199}{200}\)
C. \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A = 1-\frac{1}{2}+\frac{1}{4}+..+\frac{1}{512}\)
\(A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)
Các thành viên team mau chóng làm bài test của Phó team nữa nhé !
Chào chủ team FA muôn năm, mk vào team math is easy rùi nên không làm bài test đâu nha
thôi thôi tôi lạy bà vừa nãy đã có người làm thế rồi giờ đến lượt bà
A= 1999/2000
B = 199/200
C= 511/512
thôi thôi thôi thôi thôi thôi thôi
@ Đặng Phong
Ko làm bài test chứng tỏ bn ko bt làm rồi
@ Đức Anh
Chx chết ko phải lạy Ok
Ko làm thì mời bạn đi chỗ khác ko tiễn
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}=1-\frac{1}{2000}=\frac{1999}{2000}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{39800}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{190.200}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{190}-\frac{1}{200}=1-\frac{1}{200}=\frac{199}{200}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}+\frac{1}{512}\)
=> \(2.C=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}+\frac{1}{256}\)
=> \(2.C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}+\frac{1}{512}\right)\)
=> \(C=1-\frac{1}{512}=\frac{511}{512}\)
đầy người chả lạy kể cả khi ko chết lag
thôi thôi ko chửi nhau nữa làm thì làm ko làm thì thôi
thôi ông ơi hỏi thì hỏ cả ý
\(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\) + \(\frac{1}{3x4}\) + ....... + \(\frac{1}{1999x2000}\)
= \(1\) - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ........ + \(\frac{1}{1999}\) - \(\frac{1}{2000}\)
= \(1\) - \(\frac{1}{2000}\)
= \(\frac{1999}{2000}\)
~ Hok T ~
còn bài nào nữa ko bn
@ Thỏ con cute
Bây h mk off có chút việc tầm chiều hoặc 1 lát nx sẽ đăng bài sau ạ
Các bn team Fa muôn năm trả lời cux đc ạ
A =\(\frac{1999}{2000}\)
B=\(\frac{199}{200}\)
C= \(\frac{511}{512}\)
MK THẤY CHÌNH BÀY DÀI LẮM ,NÊN MK VIẾT MỖI KẾT QUẢ ĐC KO BN
\(\frac{1}{2}\) + \(\frac{1}{6}\) + \(\frac{1}{12}\) + \(\frac{1}{20}\) + ........ + \(\frac{1}{39800}\)
= \(\frac{1}{1x2}\) + \(\frac{1}{2x3}\) + \(\frac{1}{3x4}\) + \(\frac{1}{4x5}\)+ ....... + \(\frac{1}{199x200}\)
= \(1\) - \(\frac{1}{2}\) + \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{4}\) - \(\frac{1}{5}\)+ ....... + \(\frac{1}{199}\) - \(\frac{1}{200}\)
= \(1\) - \(\frac{1}{200}\)
= \(\frac{199}{200}\)
~ Hok T ~
A = 1999/2000
B = 199/200
C = 511/512
Hok T
Trả lời :
Quá dễ luôn
~HT~
\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\) + ....... + \(\frac{1}{512}\)
A = \(\frac{1x2}{2x2}\) + \(\frac{1x2}{4x2}\) + \(\frac{1x2}{8x2}\)+ ....... + \(\frac{1x2}{512x2}\)
A = \(\frac{2}{4}\)+ \(\frac{2}{8}\) + \(\frac{2}{16}\)+ ........ + \(\frac{2}{1024}\)
A x 2 = \(\frac{1}{4}\)+ \(\frac{1}{8}\) + \(\frac{1}{16}\)+ ......... + \(\frac{1}{1024}\)
A x 2 = \(\frac{1}{2x2}\) + \(\frac{1}{2x4}\) + \(\frac{2}{4x4}\)+ ........ + \(\frac{1}{32x32}\)
A x 2 = \(\frac{1}{2}\)- \(\frac{1}{2}\)+ ......... + \(\frac{1}{32}\) - \(\frac{1}{32}\)
A x 2 = \(\frac{1}{2}\)- \(\frac{1}{32}\)
A x 2 = \(\frac{15}{32}\)
A = \(\frac{15}{32}\): \(2\)
A = \(\frac{15}{64}\)
~ Chắc z ~
chịu