Thực hiện phép tính:
\(\left(\frac{1}{9}\right)^{2015}\cdot9^{2015}-96^2:24^2\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{2^{19}.\left(2^3\right)^3+15.\left(2^2\right)^9.\left(3^2\right)^4}{2^9.3^9.2^{10}+\left(2^2.3\right)^{10}}=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^{19}.3^9+2^{20}.3^{10}}=\frac{2^{18}.3^8.\left(2.3+15\right)}{2^{19}.3^9.\left(1+2.3\right)}\)
\(=\frac{2^{18}.3^8.21}{2^{19}.3^9.7}=\frac{21}{2.3.7}=\frac{1}{2}\)
Câu 1 :
\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^5}\)
= \(\frac{5^{32}.2^{86}.5^{43}}{\left(-2\right)^{87}.5^{15}}\)
= \(\frac{5^{72}.\left(-2\right)^{86}}{\left(-2\right)^{87}.5^{75}}\)
= \(\frac{1}{-2}\)
Câu 2 :
\(\frac{5^4.18^4}{125.9^5.16}\)
= \(\frac{5^4.2^4.3^8}{5^3.3^{10}.2^4}\)
= \(\frac{5}{3^2}\)
= \(\frac{5}{9}\)
Câu 3 :
\(\frac{9^{18}.2^{29}}{8^9.27^{12}}\)
= \(\frac{3^{36}.2^{29}}{2^{27}.3^{36}}\)
= \(2^2\)
= 4
\(b.\)ghi lại đề nha bn
\(=\frac{2.2306}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{230.231}{2}}}\)
\(=\frac{2.2306}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{230.231}}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{230.231}\right)}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{230}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+2.\left(\frac{1}{2}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+1-\frac{2}{231}}\)
\(=\frac{2.2306}{2-\frac{2}{231}}\)
\(=\frac{2.2306}{2\left(1-\frac{1}{231}\right)}\)
\(=\frac{2306}{1-\frac{1}{231}}\)
mình nha bn thanks nhìu <3
a) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{1}{2017}\)
\(\left(\frac{1}{9}\right)^{2015}.9^{2015}-96^2:24^2\)
\(=1-\left(4\right)^2\)
\(=1-16=-15\)