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a, 2 x 14 x 12 + 4 x 6 x 21 + 8 x 3 x 62
= 24 x 14 + 24 x 21 + 24 x 62
= 24 x ( 14 + 21 + 62 )
= 24 x 97
= 21 x ( 97 + 3 )
= 21 x 100
= 2100
b, và c, ( Tương tự câu a, )
\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
Ta có : D = (1 + 2 + 22 + 23 + ....... + 22004) - 22005
Đặt A = 1 + 2 + 22 + 23 + ....... + 22004
=> 2A = 2 + 22 + 23 + ....... + 22005
=> 2A - A = 22005 - 1
=> A = 22005 - 1
Thay vào ta có : D = (1 + 2 + 22 + 23 + ....... + 22004) - 22005
=> D = 22005 - 1 - 22005
=> D = -1
a,\(A=1993^{1^{2\times3\times4\times...\times1994}}=1993^1=1993\)
b,\(B=1994^{\left(225-1^2\right)\times\left(225-2^2\right).....\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\left(225-15^2\right)...\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\left(225-225\right)...\left(225-50^2\right)}\)
\(=1994^{\left(225-1^2\right)\times\left(225-2^2\right)...\times0\times...\left(225-50^2\right)}\)
\(=1994^0=1\)
c, \(C=\frac{2^{10}\times3^{31}+2^{40}\times3^6}{2^{11}\times3^{31}+2^{41}\times3^6}\)
\(=\frac{2^{10}\times3^6\times\left(1\times3^{25}+2^{30}\times1\right)}{2^{11}\times3^6\times\left(1\times3^{25}+2^{30}\times1\right)}\)
\(=\frac{2^{10}}{2^{11}}=\frac{1}{2}\)
\(A=1+2+2^2+2^3+2^4+...+2^{49}\)
\(2A=2+2^2+2^3+2^4+..+2^{50}\)
\(2A-A=\left(2+2^2+2^3+2^2+...+2^{50}\right)-\left(1+2+2^2+2^3+...+2^{49}\right)\)
\(\Rightarrow A=2^{50}-1\)