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\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
Nhận xét:
\(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{1}{10}\)
\(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{45}+\frac{1}{45}+\frac{1}{45}=\frac{1}{15}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)
Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\) (Đpcm)
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
\(=\frac{3.\left(\frac{1}{51}-\frac{4}{57}+\frac{11}{59}\right)}{5.\left(\frac{1}{51}-\frac{4}{57}+\frac{11}{59}\right)}=\frac{3.1}{5.1}=\frac{3}{5}\)
Ta có:\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+............+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+.........+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+........+\frac{1}{100}\right)\)
\(>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{1}{2}\)
\(\left(\frac{1}{51}+\frac{1}{52}+..........+\frac{1}{75}\right)+\left(\frac{1}{76}+........+\frac{1}{100}\right)\)
\(< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}< 1\)
\(\Rightarrowđpcm\)
Câu hỏi của Quỳnh Anh - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo câu 1 2 cách 2 bạn hướng dẫn nhé!