1. Tinh a \(\left(6^9.2^{10}+12^{10}\right)+\left(2^{19}.27^3+15.4^9.9^4\right)\)

2. So sanh A va B.

a) \(A=\frac{-2012}{4025};B=\frac{-1999}{3997}\)

b) \(A=3^{21};B=2^{31}\)

c) \(A=\frac{2011}{1.2}+\frac{2011}{3.4}+\frac{2011}{5.6}+....+\frac{2011}{1999.2000};\)\(B=\frac{2012}{1001}+\frac{2012}{1002}+\frac{2012}{1003}+....+\frac{2012}{2000}\)

a)A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)                b)B=\(\frac{45}{19}-\left(\frac{1}{2}\left(\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right)^-\right)^{-1}\)     c)C=\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}\)

d)D=\(\frac{2^{21}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)                e) E=\(\left(6^9.2^{10}+12^{10}\right):\left(2^{19}.27^3+15.4^9.9^4\right)\)

f) F=\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.24^3+45^6}\)          g)G=\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)         h)H=\(x+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}\)với x=-1/3

1 . Tinh : a   , \(\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{14}\right).....\left(1-\frac{1}{5050}\right)\)b,\(\frac{^{2^{19}}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}\)c,\(\frac{18.\frac{19}{2}.\frac{20}{3}.\frac{21}{4}.....\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}.....\frac{36}{17}}\)giup mjk nha mjk tjk cho

a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)

b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)

c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)

Tính:

a.A =  \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)

b. B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)

c. C = \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)