NT
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
PA
10 tháng 7 2016
a.
\(4^{10}\times8^{15}=\left(2^2\right)^{10}\times\left(2^3\right)^{15}=2^{20}\times2^{45}=2^{65}\)
b.
\(4^{15}\times5^{30}=\left(2^2\right)^{15}\times5^{30}=2^{30}\times5^{30}=\left(2\times5\right)^{30}=10^{30}\)
c.
\(\frac{27^{16}}{9^{10}}=\frac{\left(3^3\right)^{16}}{\left(3^2\right)^{10}}=\frac{3^{48}}{3^{20}}=3^{28}\)
d.
\(A=\frac{72^3\times54^2}{108^4}=\frac{\left(8\times9\right)^3\times\left(27\times2\right)^2}{\left(27\times4\right)^4}=\frac{\left(2^3\times3^2\right)^3\times\left(3^3\times2\right)^2}{\left(3^3\times2^2\right)^4}=\frac{2^9\times3^6\times3^6\times2^2}{3^{12}\times2^8}=2^3=8\)
e.
\(B=\frac{3^{10}\times11+3^{10}\times5}{3^9\times2^4}=\frac{3^{10}\times\left(11+5\right)}{3^9\times16}=\frac{3\times16}{16}=3\)
Chúc bạn học tốt
DH
21 tháng 6 2017
a, \(A=\dfrac{3^{10}.11+3^{10}.5}{3^9.2^4}=\dfrac{3^{10}.\left(11+5\right)}{3^9.2^4}\)
\(=\dfrac{3^{10}.2^4}{3^9.2^4}=3\)
b, \(B=\dfrac{2^{10}.13+2^{10}.65}{2^8.104}=\dfrac{2^{10}.78}{2^8.104}\)
\(=\dfrac{2^2.3}{4}=3\)
c, \(C=\dfrac{4^9.36+64^4}{16^4.100}=\dfrac{\left(2^2\right)^9.36+\left(2^6\right)^4}{\left(2^4\right)^4.100}\)
\(=\dfrac{2^{18}.36+2^{24}}{2^{16}.100}=\dfrac{2^{18}.\left(36+2^6\right)}{2^{16}.100}\)
\(=\dfrac{2^4.100}{100}=2^4=16\)
Câu d làm tương tự! Chúc bạn học tốt!!!
\(A=\left(4^9\cdot36+64^4\right)\div\left(16^4\cdot100\right)\)
\(A=\left[4^9\cdot4\cdot9+\left(4^3\right)^4\right]\div\left[\left(4^2\right)^4\cdot25\cdot4\right]\)
\(A=\left(4^{10}\cdot9+4^{4\cdot3}\right)\div\left[4^{2\cdot4}\cdot25\cdot4\right]\)
\(A=\left(4^{10}\cdot9+4^{12}\right)\div\left(4^8\cdot25\cdot4\right)\)
\(A=\left(4^{10}\cdot9+4^{10}\cdot4^2\right)\div\left(4^8\cdot25\cdot4\right)\)
\(A=\left(4^{10}\cdot9+4^{10}\cdot16\right)\div\left(4^8\cdot25\cdot4\right)\)
\(A=4^{10}\cdot\left(9+16\right)\div\left(4^8\cdot25\cdot4\right)\)
\(A=4^{10}\cdot25\div\left(4^8\cdot25\cdot4\right)\)
\(A=\frac{4^{10}\cdot25}{4^8\cdot25\cdot4}\)
\(A=\frac{4^2\cdot1}{1\cdot1\cdot4}\)
\(A=4\)
\(B=72^3\cdot54^2\div108^4\)
Ta lần lượt phần tích \(72,54,108\) ra thừa số nguyên tố.
\(72=2^3\cdot3^2\).
\(54=2\cdot3^3\)
\(108=2^2\cdot3^3\)
\(\Rightarrow B=\left(2^3\cdot3^2\right)^3\cdot\left(2\cdot3^3\right)^2\div\left(2^2\cdot3^3\right)^4\)
\(B=\left(2^3\right)^3\cdot\left(3^2\right)^3\cdot2^2\cdot\left(3^3\right)^3\div\left[\left(2^2\right)^4\cdot\left(3^3\right)^4\right]\)
\(B=2^{3\cdot3}\cdot3^{2\cdot3}\cdot2^2\cdot3^{3\cdot2}\div\left(2^{2\cdot4}\cdot3^{3\cdot4}\right)\)
\(B=2^9\cdot3^6\cdot2^2\cdot3^6\div\left(2^8\cdot3^{12}\right)\)
\(B=\left(2^9\cdot2^2\div2^8\right)\cdot\left(3^6\cdot3^6\div3^{12}\right)\)
\(B=2^{9+2-8}\cdot3^{6+6-12}\)
\(B=2^3\cdot1\)
\(B=8\)