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A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
A = \(\left(-\frac{1.3}{2.2}\right)\left(-\frac{2.4}{3.3}\right)...\left(-\frac{2013.2015}{2014.2014}\right)\)
A = \(-\left[\frac{\left(1.2....2013\right)\left(3.4....2015\right)}{\left(2.3....2014\right)\left(2.3...2014\right)}\right]\)
A = \(-\left(\frac{2015}{2014.2}\right)\)
A = \(-\frac{2015}{4028}\)
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\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)..........\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.........\frac{100}{99}\)
\(=\frac{100}{2}=50\)
\(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).........\left(\frac{1}{100}-1\right)\)
\(=-\frac{1}{2}.-\frac{2}{3}..........-\frac{99}{100}\)
\(=\frac{-1}{100}\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5.....100}{2.3.4.....99}\)
\(=\frac{100}{2}=50\)
a) \(\left(-\frac{1}{4}\right)^0=1\)
b) \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)
c) \(\left(\frac{4}{5}\right)^{-2}=\frac{25}{16}\)
d) \(\left(0,5\right)^{-3}=8\)
e) \(\left(-1\frac{1}{3}\right)^4=\left(-\frac{4}{3}\right)^4=\frac{256}{81}\)
a, \(\left(\frac{-1}{4}\right)^0\) = 1
Bất kỳ số nguyên nào nếu có mũ bằng 0 đều bằng 1
b, \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)
\(\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1014}{1014}\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-1\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...0.\left(1-\frac{1015}{1014}\right)\)
\(=0\)