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Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=...........................\)
\(=\frac{3^{32}-1}{2}\)
\(B=3^{32-1}\)
=> \(A< B\)
1/ \(=-\frac{64}{27}.\frac{243}{32}\)
\(=-\frac{243}{16}\)
2/ \(=\frac{1}{81}.\frac{5361441}{64}\)
\(=\frac{6561}{64}\)
3/ \(=-\frac{2197}{512}.36,71356045\)
\(=-\frac{2048}{13}\)
tíc mình nha
1\(\left(-\frac{4}{3}\right)^3.\left(\frac{9}{16}\right)^5=-\frac{2187}{16384}\)
2\(\left(\frac{1}{3}\right)^4.\left(-\frac{9}{2}\right)^6=\frac{6561}{64}\)
3\(\left(-\frac{13}{8}\right)^3.\left(-\frac{32}{13}\right)^4=-41,00457607\)
\(\left(\frac{2}{5}\right)^6:\left(\frac{2}{5}\right)^4=\left(\frac{2}{5}\right)^2=\frac{4}{25}\)
\(\left(\frac{3}{16}\right)^2:\left(\frac{9}{8}\right)^2=\frac{1}{12}\)
\(\left(\frac{2}{7}-\frac{1}{2}\right)^2=\frac{9}{196}\)
\( A= 3 ( 4^2+1).(4^4+1).(4^8+1) - ( 4^{16}+1) - \frac{4^{32}}{5}\)