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a) \(\left(0,25+\frac{3}{4}:1,25+1\frac{1}{3}:2\right)\)\(=\left(\frac{1}{4}+\frac{3}{4}\right):\frac{5}{4}+\frac{4}{3}:2\)\(=\frac{4}{5}+\frac{2}{3}=\frac{22}{15}\)
b) \(2^3+3\left(\frac{-3}{2}\right)^0.\left(\frac{1}{2}\right)^2.4+\left[\left(-2\right)^2:\frac{1}{2}\right]:8\)\(=8+3.1.\frac{1}{4}.4+\left[4:\frac{1}{2}\right]:8\)
\(=8+3+8:8=\frac{19}{8}\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)( có 2013 thừa số )
\(A=\left(-\frac{3}{2^2}\right).\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right).....\left(-\frac{\text{4056196}}{2014^2}\right)\)
\(-A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{4056196}{2014^2}=\frac{1.3.2.4.3.5....2013.2015}{2.2.3.3.4.4.....2014.2014}\)
\(-A=\frac{\left(1.2.3...2013\right).\left(3.4.5.6...2015\right)}{\left(2.3.4.5....2014\right).\left(2.3.4.5...2014\right)}=\frac{1.2015}{2.2014}=\frac{2015}{4028}\)
\(A=-\frac{2015}{4028}\)
Vậy.....
-A=(\(1-\frac{1}{2^2}\)) . (\(1-\frac{1}{3^2}\))......(\(1-\frac{1}{2014^2}\))
-A= \(\frac{3}{4}\). \(\frac{8}{9}\). ...... \(\frac{4056195}{4056196}\)
-A= \(\frac{1.3.2.4.......2013.2015}{2.2.3.3.......2.14.2014}\)
-A= \(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\)
-A= \(\frac{2015}{2014.2}\)
-A=\(\frac{2015}{4028}\)
Câu một \(=3^2.\frac{1}{3^5}.\left(3^4\right)^2.\frac{1}{3^3}=3^{10}.\frac{1}{3^8}=3^2=9\)
Câu hai \(=\left(2^2.2^5\right):\left(2^3.\frac{1}{2^4}\right)=\frac{2^7}{\frac{2^3}{2^4}}=2^8=256\)
Chờ chút nhá :D
a.
\(-2^3+2^2+\left(-1\right)^{2013}=-8+4-1=-5\)
b.
\(\left(3^3\right)^2-\left[\left(-2\right)^3\right]^2-\left(-5\right)^2=27^2-\left(-8\right)^2-25=729-64-25=640\)
c.
\(2^3+3\times\left(-\frac{1}{2016}\right)^0-\left(\frac{1}{2}\right)^2\times4-\left[\left(-2\right)^2\div\frac{1}{2}\right]=8+3\times0-\frac{1}{4}\times4-\left(4\times2\right)=8+3-1-8=2\)