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\(0,0\left(8\right)=\dfrac{1}{10}.0,\left(8\right)=\dfrac{1}{10}.0,\left(1\right).8=\dfrac{1}{10}.\dfrac{1}{9}.8=\dfrac{4}{45}\)
\(0,1\left(2\right)=0,1+0,0\left(2\right)\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(2\right)=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(1\right).2\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{9}.2=\dfrac{9}{90}+\dfrac{2}{90}=\dfrac{11}{90}\)
\(0,1\left(23\right)=0,1+0,0\left(23\right)=\dfrac{1}{10}+\dfrac{1}{10}.0,23\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(01\right).23\)
\(\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{99}.23=\dfrac{99}{990}+\dfrac{23}{990}=\dfrac{122}{990}=\dfrac{61}{495}\)
a) \(\left(-\frac{5}{2}\right)^2:\left(-15\right)-\left(-0,45+\frac{3}{4}\right).\left(-1\frac{5}{9}\right)\)
= \(-\frac{25}{4}:\left(-15\right)-\left(\frac{9}{20}+\frac{15}{20}\right).\left(-\frac{14}{9}\right)\)
=\(-\frac{25}{4}.\frac{1}{-15}-\frac{6}{5}.\left(-\frac{14}{9}\right)\)
= \(\frac{-5}{12}-\frac{8}{5}\)
= \(\frac{\left(-25\right)-96}{60}\)
= \(\frac{\left(-25\right)+\left(-96\right)}{60}\)
=\(\frac{121}{60}\)
b) \(\left(\frac{-1}{3}\right)-\left(\frac{-3}{5}\right)^0+\left(1-\frac{1}{2}\right)^2:2\)
= \(\left(\frac{-1}{3}\right)-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}\)
=\(\left(\frac{-1}{3}\right)-\frac{3}{3}+\frac{1}{4}.\frac{1}{2}\)
= \(\frac{-4}{3}+\frac{1}{8}\)=\(\frac{-32+3}{24}\)
=\(\frac{-29}{24}\)
c) E=\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{3}{5}\)
d)\(\frac{5^4.20^4}{25^5.4^5}\)
=\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)
=\(\frac{100^4}{100^5}\)
=\(\frac{1}{100}\)
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(\frac{1}{2.4}+1\right).....\left(1+\frac{1}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{4}{1.3}.\frac{9}{2.4}......\frac{4064256}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{2.2}{1.3}.\frac{3.3}{2.4}.....\frac{2016.2016}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{2.3....2016}{1.2....2015}.\frac{2.3.....2016}{3.4....2017}\right)\)
\(=\frac{1}{2}.2016.\frac{2}{2017}\)
\(=1008.\frac{2}{2017}=\frac{2016}{2017}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
