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a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)
\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)
Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)
\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)
\(2A=\frac{12}{3}-\frac{12}{99}\)
\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+....+\frac{2019}{2018.2019}\)
\(=\frac{2019}{1}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\frac{2018}{2019}\)
\(=2018\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+\frac{2019}{20}+\frac{2019}{30}+\frac{2019}{2018.2019}\)
\(A=\frac{2019}{1.2}+\frac{2019}{2.3}+\frac{2019}{3.4}+\frac{2019}{4.5}+\frac{2019}{5.6}+...+\frac{2019}{2018.2019}\)
\(A=2019.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(A=2019.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=2019.\left(1-\frac{1}{2019}\right)\)\(=2019.\frac{2018}{2019}=2018\)
Vậy A = 2018
-Dấu " . " là dấu nhân.
Đặt biểu thức trên là A ta có:
A = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{48}\)+ \(\frac{1}{96}\)
A x 3 = \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)
A x 3 = \(1\)+ \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)
A x 3 = 2 - \(\frac{1}{32}\)= \(\frac{63}{32}\)
A = \(\frac{63}{32}\): 3 = \(\frac{63}{96}\)
a ) \(\frac{7}{5}< \frac{3}{2}\)
b ) \(\frac{2}{5}>\frac{3}{8}\)
c ) \(\frac{5}{12}< \frac{3}{4}\)
d ) \(\frac{8}{12}=\frac{10}{15}\)
Trả lời
\(A=\frac{11}{12+13}+\frac{12}{13+14}+\frac{1}{14+15}\)
Hay
\(A=\frac{11}{12+13}+\frac{12}{13+14}+\frac{13}{14+15}\)
mong xem lại hộ cái
\(\frac{2}{3}+\frac{5}{8}+\frac{9}{3}:\frac{4}{7}.\left(\frac{2}{6}+\frac{1}{1}\right)\)
\(=\frac{2}{3}+\frac{5}{8}+3:\frac{4}{7}.\left(\frac{1}{3}+1\right)\)
\(=\frac{2}{3}+\frac{5}{8}+\frac{21}{4}.\frac{4}{3}\)
\(=\frac{2}{3}+\frac{5}{8}+7\)
\(=\frac{16}{24}+\frac{15}{24}+7\)
\(=\frac{31}{24}+7\)
\(=\frac{19}{12}\)
Mình không biết đúng hay sai, nhưng mà kết bạn nha!
1\(\frac{5}{7}\)X \(\frac{3}{4}\)= \(\frac{9}{7}\)
\(\frac{10}{11}\): 1\(\frac{1}{3}\)= \(\frac{15}{22}\)
Ta đặt :1 = 5/7 x 3/4 = 9/7
=> : 10/11 : 1 X 1/3 = ?
<+> : 15/22 .
cHẮC CHẮN 100%%%
\(\frac{5}{12}\)+\(\frac{7142128}{12243648}\)=\(\frac{5101520}{12243648}\)+\(\frac{7142128}{12243648}\)=\(\frac{12243648}{12243648}\)=1
\(\frac{5}{12}+\frac{7142128}{12243648}\)
\(=\frac{5}{12}+\frac{7}{12}=1\)