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I don't now
sorry
...................
nha
\(A=\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{3^{15}}=9\)
\(B=\frac{45^{10}.5^{10}}{75^{10}}=\frac{5^{10}.3^{20}.5^{10}}{5^{20}.3^{10}}=3^{10}\)
\(C=\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,4\right)^5.\left(0,2\right)^5}{0,4^6}=\frac{0,2^5}{0,2^2}=0,2^3\)
\(D=\frac{8^{10}+4^{10}}{8^{11}+4^{11}}=\frac{4^{10}\left(2^{10}+1\right)}{4^{11}\left(2^{11}+1\right)}=\frac{2^{10}+1}{2^{13}+1}\)

\(\frac{15^3+5\times15^2-5^3}{18^3+6\times18^2-6^3}\) \(=\frac{5^3.3^3+5.5^2.3^2-5^3}{3^3.6^3+6.6^2.3^2-6^3}\)\(=\frac{5^3.\left(3^3+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)=\(\frac{5^3}{6^3}\)
^-^ chúc bạn học tốt

Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)

=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)