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1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
\(F=\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^9\div\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
\(F=\frac{\left(\frac{2.5}{5}\right)^7+\left(\frac{9.16}{4.3}\right)^3}{2^7.5^2+2^9}=\frac{2^7+12^3}{2^7.5^2+2^9}=\frac{2^7+2^6.3^3}{2^7.5^2+2^9}=\frac{2^6.\left(2+3^3\right)}{2^7.\left(5^2+2^2\right)}=\frac{2^6.29}{2^7.29}\)
\(F=\frac{1}{2}\)
2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)
3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
=2/7-2/7=0