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a) \(\left(-\frac{2}{3}\right)^2:\frac{1}{3}-\left|-1\frac{1}{2}\right|=\frac{4}{9}:\frac{1}{3}-\frac{3}{2}=\frac{4}{3}-\frac{3}{2}=-\frac{1}{6}\)
b) \(\left(\frac{1}{2}-\frac{3}{5}\right)^2+\frac{2}{3}\left|\frac{3}{4}-\frac{1}{2}\right|+2012^0=\left(-\frac{1}{10}\right)^2+\frac{2}{3},\frac{1}{4}+2012^0\)
\(=\frac{1}{100}+\frac{1}{6}+1=\frac{353}{300}\)
c) \(\left(3^2:\frac{1}{3}\right)+2^3+\frac{1}{2}+\frac{1}{4}-6=3^3+2^3+\frac{3}{4}-6=29\frac{3}{4}\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)
\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)
\(=10:\left(\frac{-2}{3}\right)\)
\(=-15\)
b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\frac{-8}{27}+\frac{1}{10}\)
\(=\frac{-8}{3}+\frac{1}{10}\)
\(=\frac{-77}{30}\)
c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)
\(=\frac{-1}{3}\)
\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)
\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)
\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)
\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)
\(=-15\)
\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)
\(=-\frac{8}{3}+\frac{1}{10}\)
\(=-\frac{77}{30}\)
\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)
\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)
\(=\frac{1}{3}\)
\(2^3+3.\left(\frac{2}{3}\right)^0-2+\left[\left(-2\right)^2:\frac{1}{2}\right]-8\)
đổi p/s \(\left(\frac{2}{3}\right)^0=1\)
xong tính trong ngoặc vuông,
r xử dụng tính chất phân phối
Lời giải:
\(A=(1-\frac{1}{\frac{2(2+1)}{2}})(1-\frac{1}{\frac{3(3+1)}{2}})....(1-\frac{1}{\frac{2012(2012+1)}{2}})=(1-\frac{2}{6})(1-\frac{2}{12})(1-\frac{2}{20})....(1-\frac{2}{2012.2013})\)
Xét thừa số tổng quát:
\(1-\frac{2}{n(n+1)}=\frac{n^2+n-2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}\)
Do đó:
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2011.2014}{2012.2013}\\ =\frac{1.2.3...2011}{2.3.4...2012}.\frac{4.5.6...2014}{3.4.5...2013}\\ =\frac{1}{2012}.\frac{2014}{3}=\frac{1007}{3018}\)