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A=1+2+22+23+...+263
2A=2+22+23+...+263+264
\(-\)
\(A=1+2+2^2+....+2^{63}\)
\(A=2^{64}-1\)
Vậy A=264-1
\(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}\)
\(=\frac{2^{19}.5^2.3^{11}+2^{14}.3^{10}.5^2}{2^{17}.3^{12}.5^4-2^{18}.3^{11}.5^3}\)
\(=\frac{2^{14}.3^{10}.5^2\left(2^5.3+1\right)}{2^{17}.3^{11}.5^3\left(3.5-2\right)}=\frac{97}{2^3.3.5.13}=\frac{97}{1560}\)
\(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}\)
tớ chỉ gợi ý thôi còn tự làm lấy
đặt thừa số chung rồi rút gọn
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=\(\frac{7.\left(2^2\right)^5.3^{10}.3+2^{10}.2^3.\left(3^2\right)^5}{2^{10}.3^{10}+2^{10}.3^{10}.2^2}\)
=\(\frac{7.2^{10}.3^{10}.3+2^{10}.2^3.3^{10}}{2^{10}.3^{10}+2^{10}.3^{10}.2^2}\)
=\(\frac{2^{10}.3^{10}\left(7.3+2^3\right)}{2^{10}.3^{10}\left(1+2^2\right)}\)
=\(\frac{7.3+2^3}{1+2^2}\)
\(\frac{7.4^5.3^{11}+2^{13}.9^5}{6^{10}+2^{12}.3^{10}}=\frac{7.\left(2^2\right)^5.3^{11}+2^{13}.\left(3^2\right)^5}{\left(2.3\right)^{10}+2^{12}.3^{10}}=\frac{7.2^{10}.3^{11}+2^{13}.3^{10}}{2^{10}.3^{10}+2^{12}.3^{10}}\)
Tự làm tiếp...
\(A=3^0+3^1+3^2+......+3^{2018}\)
\(3A=3.\left(3^0+3^1+3^2+.....+3^{2018}\right)\)
\(3A=3^1+3^2+3^3+........+3^{2019}\)
\(3A-A=\left(3^1+3^2+3^3+......+3^{2019}\right)-\left(3^0+3^1+3^2+.....+3^{2018}\right)\)
\(2A=3^{2019}-3^0\)
\(A=\left(3^{2019}-3^0\right):2\)
\(B=6^{10}+6^{11}+6^{12}+....+6^{2012}\)
\(6B=6.\left(6^{10}+6^{11}+6^{12}+.....+6^{2012}\right)\)
\(6B=6^{11}+6^{12}+6^{13}+.......+6^{2013}\)
\(6B-B=\left(6^{11}+6^{12}+6^{13}+......+6^{2013}\right)-\left(6^{10}+6^{11}+6^{12}+.......+6^{2012}\right)\)
\(5B=6^{2013}-6^{10}\)
\(B=\left(6^{2013}-6^{10}\right):5\)