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S
5 tháng 3 2017
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)
\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=2.\frac{50}{159}=\frac{100}{159}\)
BT
3
12 tháng 6 2020
-5/9+8/15+-2/11+-4/9/7/17
(-5/9+-4/9)+(8/15+7/15+-2/11
-1+1+-2/11
-2/11
(17/5+11/4)-22/5
123/20-22/5
123/20-88/20
35/20
8 tháng 2 2019
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 = \(\frac{11}{24}.\)
VV
2 tháng 2 2018
ʇɐɥʇ ɥuɐɹ uɐq ɔɐɔ ɐl ƃunp ıɥʇ ʎɐp uǝp ɔonp ɔop uɐq ɔɐɔ ɐl ʇǝıq ɥuıɯ ƃunɥu 'ɔonp ɔop ıoɯ ıɐl ɔonƃu ʎɐox ıɐɥd ɐʌ ɔop oɥʞ ɐl ʇɐɹ ıɥʇ ʎɐu ǝɥʇ ʇǝıʌ ɐl ʇǝıq ɥuıɯ
4 tháng 7 2016
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+2^{18}.3^9.5}{2^{19}.3^9+2^{19}.2.3^9.3}\)
\(=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9.\left(1+6\right)}\)
\(=\frac{7}{2.7}\)
\(=\frac{1}{2}\)
Ta có : A=\(\frac{2}{3.8}+\frac{2}{8.13}+...+\frac{2}{48.53}\)
= \(2.\left(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{48.53}\right)\)
= \(\frac{2}{5}.\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{48.53}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{48}-\frac{1}{53}\right)\)
=\(\frac{2}{5}\left(\frac{1}{3}-\frac{1}{53}\right)\)
=\(\frac{2}{5}.\frac{50}{159}\)
=\(\frac{20}{159}\)
Vậy A=\(\frac{20}{159}\)
\(A=\frac{2}{3.8}+\frac{2}{8.13}+...+\frac{2}{48.53}\)
\(=\frac{2}{5}\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{48.53}\right)\)
\(=\frac{2}{5}\left(\frac{8-3}{3.8}+\frac{13-8}{8.13}+...+\frac{53-48}{48.53}\right)\)
\(=\frac{2}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=\frac{2}{5}\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=\frac{20}{159}\)