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a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=2\)
b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)
c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)
\(=2\div\frac{1}{2}=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}=\frac{3}{2}\)
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)
\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=1+1\)
\(=2\)
b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)
\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)
\(=\frac{25}{81}.\frac{3}{4}\)
\(=\frac{25}{108}\)
c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)
\(=\frac{7}{8}.2+\frac{9}{8}.2\)
\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)
\(=2.\frac{16}{8}\)
\(=2.2\)
\(=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)
\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}\)
\(=\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)
= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)
= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)
= \(x^8.\frac{1}{10}.1.1.1.1\)
= \(x^8.\frac{1}{10}\)
Mk ko pik co dung ko nua
a,(11/15+4/15)+(5/7+2/7)
=1+1
=2
b,5/9x(1/2+6/4)
=5/9x2
=10/9
c,1/2:(7/8+9/8)
=1/2:2
=1
d,(17/10-7/10)+1/2
=1+1/2
=3/2
\(1\frac{2}{24}\times5\frac{2}{5}\times2\times3\frac{7}{9}\times2\)x2/17
\(=\frac{26}{24}\times\frac{27}{5}\times2\times\frac{34}{9}\times2\times\frac{2}{17}\)
\(=\frac{351}{60}\times2\times\frac{34}{9}\times2\times\frac{2}{17}\)
\(=\frac{351}{30}\times\frac{34}{9}\times2\times\frac{2}{17}\)
\(=\frac{5967}{135}\times2\times\frac{2}{17}\)
\(=\frac{11934}{135}\times\frac{2}{17}\)
\(=\frac{2652}{255}\)