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b, S = 2 + 22 + 23 + ...+29.
2S = 2(2 + 22 + 23 + ... + 29).
= 22 + 23 + 24 + ... + 210.
2S - S = (22 + 23 + 24 + ... + 210) - (2 + 22 + 23 + ... + 29).
S = 210 - 2.
Vậy S = 210 - 2.
~Chúc bn học tốt!!!~
a, \(\dfrac{-1}{5}.\dfrac{6}{7}+\dfrac{3}{7}.\dfrac{3}{5}+\dfrac{2^5.27}{3^3.64}.\)
\(=\dfrac{-1}{5}.\dfrac{3}{7}.2+\dfrac{3}{7}.\dfrac{3}{5}+\dfrac{32.27}{27.64}.\)
\(=\dfrac{3}{7}\left(\dfrac{-1}{5}.2+\dfrac{3}{5}\right)+\dfrac{2.1}{1.2}.\)
\(=\dfrac{3}{7}.\dfrac{1}{5}+1.\)
\(=\dfrac{3}{35}+1.\)
\(=\dfrac{3}{35}+\dfrac{35}{35}.\)
\(=\dfrac{38}{35}.\)
~Chúc bn học tốt!!!~
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
) Ta có: 1615⋅−514⋅5424⋅56211615⋅−514⋅5424⋅5621
=1615⋅−514⋅94⋅83=1615⋅−514⋅94⋅83
=4⋅−13⋅47⋅3=4⋅−13⋅47⋅3
=12⋅−421=−4821=−167=12⋅−421=−4821=−167
b) Ta có: 5⋅75=355=75⋅75=355=7
c) Ta có: 17⋅59+59⋅17+59⋅3717⋅59+59⋅17+59⋅37
=59(17+17+37)=59(17+17+37)
=59⋅57=2563=59⋅57=2563
d) Ta có: 4⋅11⋅34⋅91214⋅11⋅34⋅9121
=4⋅11⋅3⋅94⋅121=2711=4⋅11⋅3⋅94⋅121=2711
e) Ta có: 34⋅169−75:−212034⋅169−75:−2120
=43+43=83=43+43=83
g) Ta có: 213−13⋅[−32+(23+0,4⋅5)]213−13⋅[−32+(23+0,4⋅5)]
=73−13⋅[−32+23+2]=73−13⋅[−32+23+2]
=73−13⋅76=73−13⋅76
=73−718=4218−718=3518
Bài giải
a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}\)
\(=\left(\frac{7}{12}-\frac{5}{12}+\frac{5}{6}+\frac{1}{4}\right)-\frac{3}{7}=\left(\frac{7}{12}-\frac{5}{12}+\frac{10}{12}+\frac{3}{12}\right)-\frac{3}{7}=\frac{5}{4}-\frac{3}{7}=\frac{23}{28}\)
b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{3^{28}\cdot4}=\frac{3\cdot8}{4}=6\)
\(a,\left(10\frac{2}{9}.2\frac{3}{5}\right)-6\frac{2}{9}=\frac{1196}{45}-\frac{56}{9}=\frac{1196}{45}-\frac{280}{45}=\frac{916}{45}\)
\(b,\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}=\frac{1}{7}\left(6+\frac{2}{7}+\frac{5}{7}\right)=\frac{1}{7}.7=1\)
\(c,3.136.8+4.14.6-14.150=3264+336-2100=1500\)
\(d,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(e,\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
Muốn cho số có hai chữ số giống nhau và chia hết cho 2 thì số đó phải là một trong các số 22, 44, 66, 88. Bây giờ ta tìm trong những số này số mà chia cho 5 thì dư 3.
Đó là số 88.
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