Tính:
A=\(\frac{3-\frac{3}{20}+\frac{3}{13}-\frac{3}{2013}}{7-\frac{7}{20}+\frac{7}{13}-\frac{7}{2013}}\)
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T
0
U
20 tháng 8 2018
a, 1 - 7x = 3x - 4
=> -7x - 3x = - 4 - 1
=> - 10x = - 5
=> x = 1/2
vậy_
b, đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
20 tháng 8 2018
mk chỉ bt lm mấy phần hui à!
d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)
\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)
e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)
18 tháng 3 2017
\(x+\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{9}{2}\)
\(x+\frac{3\times\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\times\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{9}{2}\)
\(x+\frac{3}{5}=\frac{9}{2}\)
\(x=\frac{39}{10}\)
\(x=3,9\)
1 tháng 7 2018
3 + \(\frac{3}{20}\)+ \(\frac{3}{13}\) + \(\frac{3}{2013}\)
X x = \(\frac{5}{3}\)
5 + \(\frac{5}{20}\) + \(\frac{5}{13}\) + \(\frac{5}{2013}\)
3 x ( 1 + \(\frac{1}{20}\) + \(\frac{1}{13}\) + \(\frac{1}{2013}\) )
X x = \(\frac{5}{3}\)
5 x ( 1 + \(\frac{1}{20}\) +\(\frac{1}{13}\) + \(\frac{1}{2013}\) )
X x \(\frac{3}{5}\) = \(\frac{5}{3}\) => X = \(\frac{25}{9}\) vậy X = \(\frac{25}{9}\)
1 tháng 7 2018
Ta có : \(X.\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{5}{3}\)
\(\Leftrightarrow X.\frac{3\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{5}{3}\)
\(\Leftrightarrow X.\frac{3}{5}=\frac{5}{3}\Rightarrow X=\frac{5}{3}:\frac{3}{5}=\frac{5}{3}.\frac{5}{3}=\frac{25}{9}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
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)
\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} - \left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{{ - 2}}{{97}}} \right) - \frac{1}{{35}} - \frac{3}{4} + \left( {\frac{{ - 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} - \frac{2}{{97}} - \frac{1}{{35}} - \frac{3}{4} - \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} - \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} - \frac{3}{4} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} - \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} - \frac{{33}}{{44}} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \frac{35}{{35}}+ \frac{-44}{{44}}- \frac{2}{{97}}\\= 1 + \left( { - 1} \right) - \frac{2}{{97}}\\ = - \frac{2}{{97}}\end{array}\)
MX
17 tháng 6 2017
\(x\cdot\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{5}{3}\)
\(x\cdot\frac{3\cdot\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\cdot\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{5}{3}\)
\(x\cdot\frac{3}{5}=\frac{5}{3}\)
\(x=\frac{5}{3}:\frac{3}{5}\)
\(x=\frac{25}{9}\)
17 tháng 6 2017
\(x.\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{5}{3}\)
\(x.\frac{3.\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5.\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{5}{3}\)
\(x.\frac{3}{5}=\frac{5}{3}\)
\(x=\frac{5}{3}:\frac{3}{5}\)
\(x=\frac{25}{9}\)
24 tháng 7 2015
\(x.\frac{\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{5}{3}\)
\(\Rightarrow x.\frac{3\left(\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\left(\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{5}{3}\)
\(\Rightarrow x.\frac{3}{5}=\frac{5}{3}\)
\(\Rightarrow x=\frac{5}{3}:\frac{3}{5}=\frac{25}{9}\)
vậy \(x=\frac{25}{9}\)
