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a) \(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
= \(\left(\frac{12}{25}+\frac{13}{25}\right)+\frac{3}{5}\)
= \(\frac{25}{25}+\frac{3}{5}\)
= \(1+\frac{3}{5}\)
= \(\frac{1}{1}+\frac{3}{5}\)
= \(\frac{5}{5}+\frac{3}{5}\)
= \(\frac{8}{5}\)
bài này tớ mà làm sai thì tớ ......................ko lm người........-_-
a) \(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
= \(\frac{12}{25}+\frac{15}{25}+\frac{13}{25}\)
= \(\frac{12+15+13}{25}\)
= \(\frac{40}{25}\)
= \(\frac{8}{5}\)
b) \(\frac{3}{2}+\frac{2}{3}+\frac{4}{3}\)
= \(\frac{9}{6}+\frac{4}{6}+\frac{8}{6}\)
= \(\frac{9+4+8}{6}\)
= \(\frac{21}{6}\)
= \(\frac{7}{2}\)
\(\frac{2}{4}+\frac{1}{4}=\frac{3}{4}\)\(\\ \frac{1}{3}+\frac{2}{3}=1\)\(\frac{9}{10}+\frac{1}{10}=1\)
\(1.a,\frac{5}{7}=\frac{5.9}{7.9}=\frac{45}{63};\frac{4}{9}=\frac{4.7}{9.7}=\frac{28}{63}.\)
\(b,\frac{7}{15},\frac{5}{3}=\frac{5.5}{3.5}=\frac{25}{15}\)
\(c,\frac{11}{12}=\frac{11.4}{12.4}=\frac{44}{48};\frac{7}{48}\)
\(d,\frac{3}{2}=\frac{3.3}{2.3}=\frac{9}{6};\frac{2}{3}=\frac{2.2}{3.2}=\frac{4}{6}\)
\(e,\frac{1}{3}=\frac{1.4}{3.4}=\frac{4}{12};\frac{5}{4}=\frac{5.3}{4.3}=\frac{15}{12};\frac{10}{12}\)
\(a,\frac{2001}{2002}.\frac{5}{7}.\frac{2002}{5}.\frac{7}{2001}=\left(\frac{2001}{2002}.\frac{7}{2001}\right).\left(\frac{5}{7}.\frac{2002}{5}\right)\)
\(=\frac{7}{2002}.\frac{2002}{7}=1\)
\(b,\frac{5}{7}.\frac{7}{9}.\frac{9}{11}.\frac{11}{13}=\left(\frac{5}{7}.\frac{7}{9}\right).\left(\frac{9}{11}.\frac{11}{13}\right)=\frac{5}{9}.\frac{9}{13}\)
\(=\frac{5}{13}\)
\(\frac{3}{6.8}+\frac{3}{8.10}+.......+\frac{3}{198.200}\)
\(=\frac{3}{2}.\left(\frac{2}{6.8}+\frac{2}{8.10}+........+\frac{2}{198.200}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+........+\frac{1}{198}-\frac{1}{200}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{200}\right)\)
\(=\frac{3}{2}.\frac{97}{600}=\frac{97}{400}\)
\(3.\left(\frac{2}{6.8}+\frac{2}{8.10}+....+\frac{2}{198.200}\right).\frac{1}{2}\)
=\(3.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{198}{200}\right).\frac{1}{2}\)
=\(3.\left(\frac{1}{6}-\frac{1}{200}\right).\frac{1}{2}\)
=.\(3.\frac{97}{600}.\frac{1}{2}\)=97/400
\(x-\frac{1}{3}=\frac{9}{20}\)
\(x=\frac{9}{20}+\frac{1}{3}\)
\(x=\frac{9\times3+1\times20}{60}\)
\(x=\frac{47}{60}\)
4 + \(\frac{3}{13}=\frac{4}{1}\)+ \(\frac{3}{13}=\frac{55}{13}\)
\(4+\frac{3}{13}=\frac{4}{1}+\frac{3}{13}=\frac{52}{13}+\frac{3}{13}=\frac{55}{13}\)
HT