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TL
A \(\frac{1}{4}+\frac{1}{7}=\frac{1\times7}{4\times7}+\frac{1\times4}{7\times4}=\frac{7}{24}+\frac{4}{28}=\frac{11}{28}\)
B\(\frac{1}{3}+\frac{1}{6}=\frac{1\times2}{3\times2}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\)
C\(\frac{1}{4}+\frac{1}{5}=\frac{1\times5}{4\times5}+\frac{1\times4}{5\times4}=\frac{5}{20}+\frac{4}{20}=\frac{9}{20}\)
\(\frac{11}{28},\frac{1}{2},\frac{9}{20}\)MSC 140
\(\frac{11}{28}=\frac{11\times5}{28\times5}=\frac{55}{140}\)
\(\frac{1}{2}=\frac{1\times70}{2\times70}=\frac{70}{140}\)
\(\frac{9}{20}=\frac{9\times7}{20\times7}=\frac{63}{140}\)
\(\frac{55}{140}< \frac{63}{140}< \frac{70}{140}\)hoặc \(\frac{1}{4}+\frac{1}{7}< \frac{1}{4}+\frac{1}{5}< \frac{1}{3}+\frac{1}{6}\)
Vậy cho B
HT
Bài giải
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b, \(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
Ta có :
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b,
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
\(\frac{2}{3}\)+ \(\frac{1}{2}\)= \(\frac{7}{6}\)
1 +\(\frac{3}{4}\)=\(\frac{7}{4}\)
\(\frac{5}{15}\)+\(\frac{4}{6}\)=1
nha bn
HT