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Đặt \(A=\frac{9+\frac{9}{11}+\frac{18}{23}-\frac{27}{37}}{8+\frac{8}{11}+\frac{16}{23}-\frac{24}{37}}-\frac{2+\frac{16}{29}-\frac{24}{13}-\frac{32}{11}}{3+\frac{24}{29}-\frac{36}{13}-\frac{48}{11}}\)\(=\frac{9\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}{8\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}-\frac{2\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}{3\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}\)
\(=\frac{9}{8}-\frac{2}{3}\)(do \(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37};1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\ne0\))
\(=\frac{27}{24}-\frac{16}{24}=\frac{11}{24}.\)
Vậy A = \(\frac{11}{24}.\)
a,\(\frac{21}{25}.\frac{11}{9}.\frac{5}{7}=\frac{21.11.5}{25.9.7}=\frac{3.7.11.5}{5^2.3^2.7}=\frac{11}{5.3}=\frac{11}{15}\)
b,\(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c, \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}=1-\frac{29}{15}=-\frac{14}{15}\)
a , \(\frac{21}{25}\times\frac{11}{9}\times\frac{5}{7}\)
\(=\frac{21\times11\times5}{25\times9\times7}\)
\(=\frac{3\times7\times11\times5}{5\times5\times3\times3\times7}\)
\(=\frac{11}{5\times3}\)
\(=\frac{11}{15}\)
b , \(\frac{5}{23}\times\frac{17}{26}+\frac{5}{23}\times\frac{9}{26}\)
\(=\frac{5}{23}\times\left(\frac{17}{26}+\frac{9}{26}\right)\)
\(=\frac{5}{23}\times\frac{26}{26}\)
\(=\frac{5}{23}\times1\)
\(=\frac{5}{23}\)
c , \(\left(\frac{3}{29}-\frac{1}{5}\right)\times\frac{29}{3}\)
\(=\frac{3}{29}\times\frac{29}{3}-\frac{1}{5}\times\frac{29}{3}\)
\(=1-\frac{29}{15}\)
ta có 1/3=10/30
1/21+1/22+...+1/30 có 10 p/số
mà 1/21>1/30
1/22>1/30
....
1/29>1/30
1/30=1/30
=>1/21+..1/30>1/30+....1/30 có 10 phân số
=>1/21+...1/30>1/3