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a,\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)=\(\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{6^9.2^{10}+6^{10}.2^{10}}\)
=\(\dfrac{2^{19}.3^9+2^{18}.3^9.5}{6^9.2^{10}.\left(1+6\right)}\)=\(\dfrac{2^{18}.3^9.\left(2+5\right)}{6^9.2^{10}.7}\)=\(\dfrac{2^{18}.3^9}{6^9.2^{10}}=\dfrac{2^{10}.2^8.3^9}{2^9.3^9.2^{10}}=\dfrac{2^8}{2^8.2}=\dfrac{1}{2}\)
b, \(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{16}}{\dfrac{-23}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-29}{16}}{\dfrac{-29}{16}}=1\)
\(=\dfrac{2^{19}.\left(3^3\right)^3-3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\dfrac{2^{19}.3^9-5.2^{18}.3^9}{2^{19}.3^9+2^{20}.3^{10}}=\dfrac{2^{18}.3^9\left(2-5\right)}{2^{19}.3^9\left(1+6\right)}=\dfrac{-3}{2.7}=-\dfrac{3}{14}\)
\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\\ =\dfrac{2^{19}.3^9+5.2^{18}.3^9}{2^9.3^9+2^{20}.3^{10}}\\ =\dfrac{2^{18}.3^9\left(2+5\right)}{2^9.3^9\left(2^{11}.3+1\right)}\\ =\dfrac{2^9.7}{2^9.12+1}=\dfrac{7}{13}\)
\(A=4,8.\left(3,1-1,5\right)+1,5.\left(4,8-3,1\right)\)
\(A=4,8.3,1-4,8.1,5+1,5.4,8-1,5.3,1\)
\(A=3,1.\left(4,8-1,5\right)-4,8\left(1,5+1,5\right)\)
\(A=3,1.3,3-4,8.3\)
\(A=10,23-14,4=-4,17\)
\(B=\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{2.3^9.2^{10}+3^{10}.\left(2^2\right)^{10}}=\dfrac{2^{19}.3^9+3^9.2^{18}.5}{2^{11}.3^9+3^{10}.2^{20}}=\dfrac{2^{18}.3^9\left(2+5\right)}{2^{11}.3^9\left(1+3.2^9\right)}=\dfrac{2^7.7}{1+3.2^9}\)
\(\dfrac{2^{19}+27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\dfrac{2^{19}+\left(3^3\right)^3+5.3.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.4\right)^{10}}\)
\(=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.4^{10}}\)
\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.\left(2^2\right)^{10}}\)
\(=\dfrac{2^{18}.3^9.\left(2.5\right)}{3^9.2^{19}+3^{10}.2^{20}}\)
\(=\dfrac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+3.2\right)}\)
\(=\dfrac{7}{2\left(1+6\right)}\)
\(=\dfrac{7}{2.7}\)
\(=\dfrac{1}{2}\)
a) \(5^{20}và2550^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}< 2550^{10}\)
=> \(5^{20}< 2550^{10}\)
b) \(999^{10}và999999^5\)
\(999^{10}=\left(999^2\right)^5=1998^5< 999999^5\)
=> \(999^{10}< 999999^5\)
c) \(\left(\dfrac{-1^{300}}{5}\right)và\left(\dfrac{-1^{500}}{3}\right)\)
\(\left(\dfrac{-1^{300}}{5}\right)=\dfrac{-1}{5}\)
\(\left(\dfrac{-1^{500}}{3}\right)=\dfrac{-1}{3}\)
\(\dfrac{-1}{5}=\dfrac{-3}{15}\)
\(\dfrac{-1}{3}=\dfrac{-5}{15}\)
=> \(\dfrac{-3}{15}>\dfrac{-5}{15}\)
=> \(\left(\dfrac{-1^{300}}{5}\right)>\left(\dfrac{-1^{500}}{3}\right)\)
\(=\dfrac{2^{19}\cdot3^9+2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{2^{18}\cdot3^9\left(5+2\right)}{2^{19}\cdot3^9\left(1+2\cdot3\right)}=\dfrac{1}{2}\)
\(=\dfrac{2\cdot3^9+3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot3^9\cdot2^{10}+2^{20}\cdot3^{10}}=\dfrac{2\cdot3^9+3^9\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}\\ =\dfrac{2\cdot3^9\left(1+2^{17}\cdot5\right)}{2^{19}\cdot3^9\left(1+2\cdot3\right)}=\dfrac{1+2^{17}\cdot5}{2^{18}\cdot7}\)
may mà m đăng để t đỡ phải viết (mỏi tay) Nguyen Ngoc Linh
\(A=\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+2^{12}}\)
\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{2^9.3^{19}+2^{12}}=\dfrac{2^{10}+5.2^8}{3^{10}+2^3}=\dfrac{2^7+5.2^5}{3^{10}}\)