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\(\frac{45^{10}.5^{10}}{75^{40}}=\frac{\left(5.3^2\right)^{10}.5^{10}}{\left(5^2.3\right)^{40}}=\frac{5^{10}.3^{20}.5^{10}}{5^{80}.3^{40}}=\frac{1}{5^{60}.3^{20}}\)
\(\frac{21^2\cdot14\cdot125}{35^2\cdot6}=\frac{3^2\cdot7^2\cdot2\cdot25\cdot3}{5^2\cdot7^2\cdot2\cdot3}=\frac{3^4\cdot7^2\cdot2\cdot5^2}{5^2\cdot7^2\cdot2\cdot3}=3^3=27..\)
\(\frac{45^3\cdot20^4\cdot18^2}{180^5}=\frac{5^3\cdot9^3\cdot4^4\cdot5^4\cdot2^2\cdot9^2}{2^{10}\cdot3^{10}\cdot5^5}=\frac{5^7\cdot9^5\cdot4^4}{4^5\cdot9^5\cdot5^5}=\frac{1}{4}=0.25\)
1/2D=1/2(1/6+1/10+......+1/45)
1/2D=1/12+1/20+1/30+.....+1/90
1/2D=1/3.4+1/4.5+1/5.6+......+1/9.10
1/2D=1/3-1/4+1/4-1/5+1/5-1/6+....+1/9-1/10
1/2D=1/3-1/10
1/2D=7/30
D=7/30:1/2
D=7/15
Ta có:\(D=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(=\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}\)
\(=2.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{10}\right)=2.\frac{7}{30}=\frac{7}{15}\)
Vậy \(D=\frac{7}{15}\)
\(\frac{\left(45.5\right)^{10}}{75}=\frac{225^{10}}{75}=3^{10}\)
\(C=\frac{45^{10}.5^{10}}{75}=\frac{225^{10}}{75}=\frac{225.225^9}{75}=\frac{75.3.225^9}{75}=3.225^9\)