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Đặt \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(\Rightarrow\frac{1}{2^2}.A=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
\(\Rightarrow\frac{1}{4}.A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}+\frac{1}{16^2}\)
\(\Rightarrow A-\frac{1}{4}.A=\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)-\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}+\frac{1}{16^2}\right)\)
\(\Rightarrow\frac{3}{4}.A=\frac{1}{2^2}-\frac{1}{16^2}=\frac{1}{4}-\frac{1}{256}=\frac{63}{256}\)
\(\Rightarrow A=\frac{63}{256}:\frac{3}{4}=\frac{21}{64}\)
K rút họn đc đâu bạn. Bạn muốn chứng minh tổng trên bé hơn hoặc lớn hơn số nào thì đc
a) \(\frac{28}{36}=\frac{7}{9}\)
b)\(-\frac{63}{90}=-\frac{7}{10}\)
c)\(\frac{40}{-120}=-\frac{1}{3}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{899}{900}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{29.31}{30.30}\)
\(=\frac{1.2.3....29}{2.3.4....30}.\frac{3.4.5....31}{2.3.4....30}\)
\(=\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
\(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3.\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4.\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
a, 11 1/4-(2 5/7+5 1/4)
= 45/4-(19/7+21/4)
= 45/4-223/28
=23/7
b, (8 5/11+3 5/8)-3 5/11
=(93/11+29/8)-38/11
=1063/88-38/11
=69/8
a, =\(11\frac{1}{4}-2\frac{5}{7}-5\frac{1}{4}\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=\frac{23}{7}\)
b, \(=8\frac{5}{11}+3\frac{5}{8}-3\frac{5}{11}\)
\(=\left(8\frac{5}{11}-3\frac{5}{11}\right)+3\frac{5}{8}\)
\(=5+3\frac{5}{8}\)
\(=\frac{69}{8}\)
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}A=1-\frac{1}{10}\)
\(\frac{1}{2}A=\frac{9}{10}\)
\(A=\frac{9}{10}:\frac{1}{2}\)
\(A=\frac{18}{10}=\frac{9}{5}\)
1) \(\frac{3^{10}+6^2}{5\cdot3^8+20}=\frac{3^{10}+3^2\cdot2^2}{5\cdot3^8+5\cdot2^2}=\frac{3^2\left(3^8+2^2\right)}{5\left(3^8+2^2\right)}=\frac{9}{5}\)
2) \(\frac{28^{15}\cdot3^{17}}{84^{16}}=\frac{28^{15}\cdot3^{17}}{28^{16}\cdot3^{16}}=\frac{3}{28}\)