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20 tháng 10 2020

\(\left(1+\frac{1}{1}\right).\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).\left(1+\frac{1}{5}\right).\left(1+\frac{1}{6}\right)\)

\(=2.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}.\frac{7}{6}\)

\(=\frac{2.3.4.5.6.7}{2.3.4.5.6}=7\)

HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023

\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)

22 tháng 3 2020

\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)

16 tháng 7 2016

\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)

\(P=\left(-1,1\right):\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)

\(P=\frac{11}{30}+\frac{1}{3}+\left(-\frac{1}{12}\right)\)

\(P=\frac{37}{60}\)

\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)

\(Q=\left(-0,928\right):\frac{4}{7}:\left[\left(-\frac{119}{36}\right).2\frac{2}{17}\right]\)

\(Q=\left(-1,624\right):\left(-\frac{245}{36}\right)\)

\(Q=\frac{1044}{4375}\)

24 tháng 9 2017

KHÓ QUÁ!!!!

25 tháng 9 2017

1007,851098

25 tháng 9 2017

\(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\left(\frac{1}{2}\right)^2\right)^6\cdot2017}{\left(\frac{1}{2}\right)^2\cdot\frac{1}{3}\cdot\left(\frac{1}{2}\right)^{13}}\)

\(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\frac{1}{2}\right)^{12}\cdot2017}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)

=\(\frac{\left(\frac{1}{2}\right)^2\cdot\left(2018-2017\right)\cdot\left(\frac{1}{2}\right)^{10}}{\left(\frac{1}{2}\right)^{15}.\frac{1}{3}}\)

\(\frac{\left(\frac{1}{2}\right)^2\cdot1\cdot\left(\frac{1}{2}\right)^{10}}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)

\(\frac{\left(\frac{1}{2}\right)^{12}}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)

\(\frac{1}{\left(\frac{1}{2}\right)^3\cdot\frac{1}{3}}\)

\(\frac{1}{\frac{1}{24}}\)

25 tháng 9 2017

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A =  - 1\end{array}\)

b)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)

\(P=\left(\dfrac{-1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{6}:2\)

\(=\left(\dfrac{1}{2}+\dfrac{3}{5}\right):3+\dfrac{1}{3}-\dfrac{1}{12}\)

\(=\dfrac{11}{10}\cdot\dfrac{1}{3}+\dfrac{1}{4}\)

\(=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{22}{60}+\dfrac{15}{60}=\dfrac{37}{60}\)

\(Q=\left(\dfrac{2}{25}-\dfrac{126}{125}\right)\cdot\dfrac{7}{4}:\left[\dfrac{-119}{36}\cdot\dfrac{36}{17}\right]\)

\(=\dfrac{-116}{125}\cdot\dfrac{7}{4}:\left(-7\right)\)

\(=\dfrac{116}{125}\cdot\dfrac{7}{4}\cdot\dfrac{1}{7}=\dfrac{29}{125}\)