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M=\(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{4^{20}\left(2^{20}+1\right)}{4^{25}+4^{15}}=\dfrac{4^{20}\left(2^{20}+1\right)}{4^{15}\left(4^{10}+1\right)}=\dfrac{2^{20}+1}{4^{10}+1}\)
T=\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{9^{10}.5^{30}}{25^{15}.3^{15}}=\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
a)\(\dfrac{6^6+6^3.3^3+3^6}{-73}\)\(=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6.1}{-73}\)
\(=\dfrac{3^6.\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6\left(64+8+1\right)}{-73}=^{ }\)\(\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=-3^6\)
b)\(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}.\left(2^{20}+1\right)}{2^{30}.\left(2^{20}+1\right)}=\dfrac{2^{40}}{2^{30}}=2^{10}=1024\)
c)\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}=\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
\(\dfrac{\text{45^{10^{ }}}.5^{10}}{75^{10}}=\dfrac{9^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}=\dfrac{9^{10}}{3^{10}}=3^{10}\)
\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)
A=6^6+6^3+3^3+3^6/-73=2^6.3^6+2^3.3^3.3^3+3^6/-73=2^6.3^6+2^3.3^6+3^6/-73=(2^6+2^3+1).3^6/-73=73.3^6/-73=-(3^6)=...
\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)
\(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{2^3.3^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}\) =\(2^3.8^5\)
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)
a, \(4^3.5^3=\left(4.5\right)^3=20^3=8000\)
b, \(6^3.5^3=\left(6.5\right)^3=30^3=27000\)
c, \(8^2.5^2=\left(8.5\right)^2=40^2=1600\)
d, \(125^3.8^3=\left(125.8\right)^3=1000^3\)
e, \(5^2.6^2.3^2=\left(5.6.3\right)^2=90^2\)
\(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{3^6.2^6+3^3.2^3.3^3+3^6}{-73}=\dfrac{3^6.64+3^6.8+3^6}{-73}=\dfrac{3^6.\left(64+8+1\right)}{-73}=-3^6=-729\)
Giải:
\(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}\)
\(=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}\)
\(=\dfrac{3^6\left(64+8+1\right)}{-73}=\dfrac{3^6.73}{-73}=-3^6=-729.\)
Vậy.....
\(N=\dfrac{3^6\cdot2^6+3^6\cdot2^3+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=-3^6\)
\(T=\dfrac{5^{20}\cdot3^{20}}{5^{30}\cdot3^{15}}=\dfrac{3^5}{5^{10}}\)