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Ta thấy \(10^{50}>10^{50}-3\)
\(\Rightarrow B=\frac{10^{50}}{10^{50}-3}>\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)
Vậy \(A< B\)
Ta có : A = 1.2 + 2.3 + 3.4 + ..... + 49.50
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 49.50.51
=> 3A = 49.50.51
= >A = 49.50.51/3 = 41650
\(10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(4^{50}=\left(4^5\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}\)nên \(10^{30}< 4^{50}\)
M = \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\)
=> 5M = 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)
=> 5M - M = ( 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)) - ( \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\))
4M = 1 - \(\left(\frac{1}{5}\right)^{50}\)
=> M = \(\frac{1-\left(\frac{1}{5}\right)^{50}}{4}\)< \(\frac{1}{4}\)
\(a.=1992-53+158-247-1592\)
\(=\left(1992-1592\right)-\left(53+247\right)+158\)
\(=400-300+158\)
\(=100+158\)
\(=258\)
\(b.=-85-\left(-10\right)-\left(-85\right)+\left(-50\right)\)
\(=-85-\left(-85\right)-\left(-10--50\right)\)
\(=0-40\)
\(=-40\)
50+50=50x2=100
\(50+50=100\)