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PN
6 tháng 7 2017
a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)
\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)
\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)
\(=5\dfrac{4}{23}.23\)
\(=\dfrac{119}{23}.23\)
\(=\dfrac{119}{23}\)
b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{3}{2}\)
\(=\dfrac{-2}{3}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{9}{6}\)
\(=\dfrac{5}{6}\)
c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)
\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)
\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)
\(=1-1\)
\(=0\)
d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
(đợi đã, mình chưa tìm được hướng làm...)
NN
Cho A= \(\frac{10^{2011+5}}{10^{2011}-2}\); B= \(\frac{10^{2011}}{10^{2011}-7}\). Hãy so sánh A và B
1
S
13 tháng 5 2017
\(A=\frac{10^{2011}+5}{10^{2011}-2}=\frac{10^{2011}-2+7}{10^{2011}-2}=1+\frac{7}{10^{2011}-2}\)
\(B=\frac{10^{2011}}{10^{2011}-7}=\frac{10^{2011}-7+7}{10^{2011}-7}=1+\frac{7}{10^{2011}-7}\)
Vì \(\frac{7}{10^{2011}-2}< \frac{7}{10^{2011}-7}\Rightarrow1+\frac{7}{10^{2011}-2}< 1+\frac{7}{10^{2011}-7}\Rightarrow A< B\)
\(2011^{2011}.\left(7^{10}:7^8-3.2^4-2^{2011}:2^{2011}\right)\)
\(=2011^{2011}.\left(7^{10-8}-3.2^4-2^{2011-2011}\right)\)
\(=2011^{2011}.\left(7^2-3.2^4-2^0\right)\)
\(=2011^{2011}.\left(49-3.16-1\right)\)
\(=2011^{2011}.\left(49-48-1\right)\)
\(=2011^{2011}.\left(1-1\right)\)
\(=2011^{2011}.0\)
\(=0\)