\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+....+\frac{1}{160^2}\)
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a) \(14:\frac{0,4x+0,6}{x}=7\)
\(\frac{0,4x+0,6}{x}=2\)
0,4x + 0,6 = 2.x
2x - 0,4x = 0,6
1,6x = 0,6
x = 0,375
b) \(\left(160\%+\frac{2}{3}x-x\right).12=660\)
\(\left(160\%+\frac{2}{3}x-x\right)=55\)
\(x\left(\frac{2}{3}-1\right)=53,4\)
\(-\frac{1}{3}x=\frac{267}{5}\)
\(x=\frac{267}{5}.\frac{3}{-1}\)
\(x=-160,2\)
c) \(1:\frac{1.2.3.4.....31}{2.2.2.3.2.4.....2.32}=2^x\)
\(1:\frac{1.2.3.4.....31}{2^{31}.2.3.4.....31.2^5}=2^x\)
\(1:\frac{1}{2^{36}}=2^x\)
\(2^{36}=2^x\)
\(x=36\)
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)
Tách 9=1+1+...+1 ( có 9 số 1)
\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )
Vậy \(A:B=10\)
a) $\frac{1}{6} + \frac{3}{2} + \frac{1}{2} = \frac{1}{6} + \left( {\frac{3}{2} + \frac{1}{2}} \right) = \frac{1}{6} + \frac{4}{2} = \frac{1}{6} + \frac{{12}}{6} = \frac{{13}}{6}$
b) $\frac{3}{8} + \frac{1}{2} + \frac{1}{8} = \left( {\frac{3}{8} + \frac{1}{8}} \right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
c) $\frac{2}{5} + \frac{6}{{10}} + \frac{3}{5} = \frac{2}{5} + \frac{3}{5} + \frac{3}{5} = \frac{{2 + 3 + 3}}{5} = \frac{8}{5}$