\(\dfrac{3^{10}.11+9^5.5}{27^3.2^4}.x=-9\)
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22 tháng 7 2023
a) \(\dfrac{27^3\cdot11+9^5\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{3^9\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{3^9\cdot\left(11+3\cdot5\right)}{3^9\cdot2^4}\)
\(=\dfrac{11+15}{16}\)
\(=\dfrac{26}{16}\)
\(=\dfrac{13}{8}\)
b) \(\dfrac{5^8+2^2\cdot25^4+2^3\cdot125^3-15^4\cdot5^4}{4^2\cdot625^2}\)
\(=\dfrac{5^8+2^2\cdot5^8+2^3\cdot5^9-3^4\cdot5^4\cdot5^4}{2^4\cdot5^8}\)
\(=\dfrac{5^8\cdot\left(1+2^2+2^3\cdot5-3^4\right)}{5^8\cdot2^4}\)
\(=\dfrac{1+4+40-81}{16}\)
\(=\dfrac{-36}{16}\)
\(=\dfrac{-9}{4}\)
c) \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(=\dfrac{-2}{6}\)
\(=-\dfrac{1}{3}\)
12 tháng 6 2017
\(1,\)
\(\dfrac{45^2.3^8.10^5}{5^5.3^7.18^5}\)
\(=\dfrac{3^4.5^2.3^8.2^5.5^5}{5^5.3^7.2^5.3^{10}}\)
\(=\dfrac{3^{12}.2^5.5^7}{5^5.3^{17}.2^5}\)
\(=\dfrac{1.5^2}{3^5.1}\)
\(=\dfrac{25}{243}\)
\(2,\)
\(\dfrac{4^5.9^4+2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{2^{10}.3^8+2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8+2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.3^8.4}{2^{10}.3^8.6}\)
\(=\dfrac{2^{12}.3^8}{2^{11}.3^9}\)
\(=\dfrac{2}{3}\)
\(3,\)
\(\dfrac{15.3^{11}+4.27^4}{9^7}\)
\(=\dfrac{3.5.3^{11}+2^2.3^{12}}{3^{14}}\)
\(=\dfrac{5.3^{12}+2^2.3^{12}}{3^{14}}\)
\(=\dfrac{3^{12}\left(5+2^2\right)}{3^{14}}\)
\(=\dfrac{3^{12}.9}{3^{14}}\)
\(=\dfrac{3^{14}}{3^{14}}\)
\(=1\)
\(4,\)
\(\dfrac{4^7.2^8}{3.2^{15}.16^2-5^2\left(2^{10}\right)^2}\)
\(=\dfrac{2^{22}}{3.2^{23}-5^2.2^{20}}\)
\(=\dfrac{2^{22}}{2^{20}.\left(-1\right)}\)
\(=\dfrac{2^{22}}{-2^{20}}\)
\(=-4\)
* Mấy bài còn lại tương tự đấy bạn tự làm đi
Mình mỏi tay lắm rồi
12 tháng 6 2017
P/s:khuyến khích tự làm,chỉ làm mẫu 1 câu:
1)\(\dfrac{45^2.3^8.10^5}{5^5.3^7.18^5}=\dfrac{\left(5.9\right)^2.3.3^7.\left(2.5\right)^5}{5^5.3^7.\left(2.9\right)^5}\)\(=\dfrac{5^2.9^2.3.3^7.2^5.5^5}{5^5.3^7.2^5.9^5}\)\(=\dfrac{5^2.9^2.3.1.1.1}{1.1.1.9^5}\)\(=\dfrac{5^2.9^2.3}{9^5}=\dfrac{5^2.9^2.3}{9^2.9^3}=\dfrac{5^2.3}{9^3}=\dfrac{75}{729}=\dfrac{25}{243}\)
CX
0
PN
3
15 tháng 7 2019
\(\frac{9^3\cdot2^{10}\cdot27^5}{4^5\cdot81^6}=\frac{3^6\cdot2^{10}\cdot3^{15}}{2^{10}\cdot3^{24}}=\frac{1}{3^3}=\frac{1}{27}\)
\(\frac{27^4\cdot2^5-3^{11}\cdot4^3}{8^2\cdot9^6}=\frac{3^{12}\cdot2^5}{2^6\cdot3^{12}}-\frac{3^{11}\cdot2^6}{2^6\cdot3^{12}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
=))
XO
15 tháng 7 2019
a) \(\frac{9^3.2^{10}.27^5}{4^5.81^6}\)= \(\frac{\left(3^2\right)^3.2^{10}.\left(3^3\right)^5}{\left(2^2\right)^5.\left(3^4\right)^5}\)= \(\frac{3^{2.3}.2^{10}.3^{3.5}}{2^{2.5}.3^{4.5}}\)= \(\frac{3^6.2^{10}.3^{15}}{2^{10}.3^{20}}\)= \(\frac{3^{21}.2^{10}}{2^{10}.3^{20}}\)= \(\frac{3^{20}.2^{10}.3}{2^{10}.3^{20}}\)= \(3\)
MV
29 tháng 10 2017
\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
28 tháng 8 2021
a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
27 tháng 5 2022
Câu 2:
a: a=2007 nên a+1=2008
\(M=a^{11}-a^{10}\left(a+1\right)+a^9\left(a+1\right)-...-a^2\left(a+1\right)+a\left(a+1\right)\)
\(=a^{11}-a^{11}-a^{10}+a^{10}+a^9-...-a^3-a^2+a^2+a\)
=a=2007
b: a=2004 nên a-1=2003
\(N=a^{11}-a^{10}\left(a-1\right)-a^9\left(a-1\right)-...-a\left(a-1\right)-1004\)
\(=a^{11}-a^{11}+a^{10}-a^{10}+a^9-...-a^2+a-1004\)
=a-1004=1000
\(\dfrac{3^{10}\cdot11+9^5\cdot5}{27^3\cdot2^4}\cdot x=-9\)
\(\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\cdot x=-9\)
\(\dfrac{3\cdot3^9\cdot16}{3^9\cdot16}\cdot x=-9\)
3x = -9
\(x=-\dfrac{9}{3}=-3\)
\(\dfrac{3\cdot3^9\cdot\left(11+5\right)}{3^9\cdot16}\cdot x=-9\)