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\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}\)
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AL
2
1 tháng 12 2015
Ta có
\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}=\frac{\left(3^2\right)^{15}.\left(5^2\right)^{43}}{\left(3^3\right)^{10}.5^{85}}=\frac{3^{30}.5^{86}}{3^{30}.5^{85}}=5\)
nhớ tick mình nha
1 tháng 12 2015
\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}=\frac{\left(3^2\right)^{15}.\left(5^2\right)^{43}}{\left(3^3\right)^{10}.5^{85}}=\frac{3^{30}.5^{86}}{3^{30}.5^{85}}=\frac{5\left(3^{30}.5^{85}\right)}{5^{30}.5^{85}}=5\)
9 tháng 1 2020
Đặt A=\(\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
\(\frac{A}{2}=\frac{13}{42}-\frac{15}{56}+\frac{17}{72}-...+\frac{197}{9702}-\frac{199}{4950}\)
\(=\frac{6+7}{6.7}-\frac{7+8}{7.8}+\frac{8+9}{8.9}-...+\frac{98+99}{98.99}-\frac{99+100}{99.100}\)
\(=\frac{1}{7}+\frac{1}{6}-\frac{1}{8}-\frac{1}{7}+\frac{1}{9}+\frac{1}{8}-...+\frac{1}{99}+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}\)
\(=\frac{1}{6}-\frac{1}{100}=\frac{47}{300}\)
\(\Rightarrow A=\frac{47}{300}.2=\frac{47}{150}\)
\(\Rightarrow Q=\frac{85}{25}+\frac{9}{10}-\frac{11}{5}+\frac{47}{150}=\frac{181}{75}\)
Vậy Q=\(\frac{181}{75}\).
11 tháng 8 2021
a) \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=-\dfrac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=\dfrac{-3}{5}\)
b: \(\dfrac{4^{30}\cdot3^{43}}{2^{57}\cdot27^{15}}\)
\(=\dfrac{2^{60}\cdot3^{43}}{2^{57}\cdot3^{45}}\)
\(=8\cdot\dfrac{1}{9}=\dfrac{8}{9}\)
PQ
3
TN
23 tháng 7 2016
\(G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(=\frac{1}{25}-\frac{4}{27}+\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}+\frac{5}{43}-\frac{4}{7}\)
\(=\left(\frac{1}{25}+\frac{-1}{25}\right)+\left(\frac{-4}{27}+\frac{-23}{27}\right)+\left(\frac{-5}{43}+\frac{5}{43}\right)-\frac{4}{7}\)
\(=0+\left(-1\right)+0-\frac{4}{7}\)
\(=\frac{-11}{7}\)
\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}=\frac{\left(3^2\right)^{15}.\left(5^2\right)^{43}}{\left(3^3\right)^{10}.5^{85}}=\frac{3^{30}.5^{86}}{3^{30}.5^{85}}=5\)
\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}=\frac{\left(3^2\right)^{15}.\left(5^2\right)^{43}}{\left(3^3\right)^{10}.5^{85}}=\frac{3^{30}.5^{86}}{3^{30}.5^{85}}=1.5=5.\)
Chúc bạn học tốt!