Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{10}{9}.\frac{11}{10}.....\frac{2006}{2005}=\frac{2006}{9}\)
Ta có:
\(\left(\frac{1}{9}+1\right).\left(\frac{1}{10}+1\right).....\left(\frac{1}{2005}+1\right)\)
\(=\left(\frac{1}{9}+\frac{9}{9}\right).\left(\frac{1}{10}+\frac{10}{10}\right).....\left(\frac{1}{2005}+\frac{2005}{2005}\right)\)
\(=\frac{10}{9}.\frac{11}{10}.....\frac{2006}{2005}\)
\(=\frac{2006}{9}\)
a)
\(\begin{array}{l}\frac{3}{7}.\left( { - \frac{1}{9}} \right) + \frac{3}{7}.\left( { - \frac{2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} + \frac{-2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} - \frac{6}{9}} \right)\\ = \frac{3}{7}.\frac{{ - 7}}{9} = \frac{{ - 1}}{3}\end{array}\)
b)
\(\begin{array}{l}\left( {\frac{{ - 7}}{{13}}} \right).\frac{5}{{12}} + \left( {\frac{{ - 7}}{{13}}} \right).\frac{7}{{12}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.\left( {\frac{5}{{12}} + \frac{7}{{12}}} \right) + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.1 + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 13}}{{13}}\\ = -1\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right]:\frac{5}{9} + \left( {\frac{4}{7} - \frac{1}{3}} \right):\frac{5}{9}\\ = \left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right].\frac{9}{5} + \left( {\frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left( {\frac{{ - 2}}{3} + \frac{3}{7} + \frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left[ {\left( {\frac{{ - 2}}{3} - \frac{1}{3}} \right) + \left( {\frac{3}{7} + \frac{4}{7}} \right)} \right].\frac{9}{5}\\ = \left( { - 1 + 1} \right).\frac{9}{5}\\ = 0.\frac{9}{5} = 0\end{array}\)
d)
\(\begin{array}{l}\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{2}{3}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{{10}}{{15}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 9}}{15}\\= \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 3}}{5}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{5}{9}.\frac{{ - 5}}{3}\\ = \frac{5}{9}.\left( {\frac{{ - 22}}{3} - \frac{5}{3}} \right)\\ = \frac{5}{9}.\frac{-27}{3}= \frac{5}{9}.\left( { - 9} \right) = - 5\end{array}\)
e)
\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} - \left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{{ - 2}}{{97}}} \right) - \frac{1}{{35}} - \frac{3}{4} + \left( {\frac{{ - 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} - \frac{2}{{97}} - \frac{1}{{35}} - \frac{3}{4} - \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} - \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} - \frac{3}{4} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} - \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} - \frac{{33}}{{44}} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \frac{35}{{35}}+ \frac{-44}{{44}}- \frac{2}{{97}}\\= 1 + \left( { - 1} \right) - \frac{2}{{97}}\\ = - \frac{2}{{97}}\end{array}\)
a)
\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)
b)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ = - 2 - \frac{5}{7}.\frac{7}{{15}}\\ = - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)
c)
\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)
d)
\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)
c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)
\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)
\(\frac{100}{-\frac{100}{41}}=-41\)
a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)
\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)
\(7:-\frac{1}{7}=-49\)
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right).\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}.\frac{99}{100}\)
\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{8.10}{9.9}.\frac{9.11}{10.10}\)
\(B=\frac{1.2.3...8.9}{2.3.4...9.10}.\frac{3.4.5...10.11}{2.3.4...9.10}\)
\(B=\frac{1}{10}.\frac{11}{2}\)
\(B=\frac{11}{20}>\frac{11}{21}\)
a,\(\left(\frac{1}{9}-1\right).\left(\frac{1}{10}-1\right)...\left(\frac{1}{2004}-1\right).\left(\frac{1}{2005}-1\right)\)
\(=\frac{-8}{9}.\frac{-9}{10}...\frac{-2003}{2004}.\frac{-2004}{2005}\)
\(=\frac{\left(-8\right).\left(-9\right)...\left(-2003\right).\left(-2004\right)}{9.10...2004.2005}\)
\(=\frac{-\left(8.9...2003.2004\right)}{9.10...2004.2005}\)
\(=\frac{-8}{2005}\)
b,Ta có: \(81^{10}-27^{13}-9^{21}\)
\(=\left(3^4\right)^{10}-\left(3^3\right)^{13}-\left(3^2\right)^{21}\)
\(=3^{40}-3^{39}-3^{42}\)
\(=3^{39}.3-3^{39}-3^{39}.3^3\)
\(=3^{39}.\left(3-1-3^3\right)\)
\(=3^2.3^{37}.\left(-25\right)\)
\(=3^{37}.\left(-225\right)⋮225\)
Vậy \(81^{10}-27^{13}-9^{21}⋮225\)