Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(C=\frac{\left(\frac{2}{3}\right)^3\times\left(-\frac{3}{4}\right)^2\times\left(-1\right)^5}{\left(\frac{2}{5}\right)^2\times\left(-\frac{5}{12}\right)^2}\)
\(C=\frac{\frac{2^3}{3^3}.\frac{\left(-3\right)^2}{4^2}.\left(-1\right)^5}{\frac{2^2}{5^2}.\frac{\left(-5\right)^2}{12^2}}\)
\(C=\frac{\frac{-\left(2^3.3^2\right)}{3^3.2^4}}{\frac{2^2.5^2}{5^2.2^4.3^2}}\)
\(C=\frac{\frac{-1}{3.2}}{\frac{1}{2^2.3^2}}\)
\(C=\frac{\frac{-1}{6}}{\frac{1}{36}}\)
\(C=-6\)
b) \(D=\frac{6^6+6^3\times3^3+3^6}{-73}\)
\(D=\frac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}\)
\(D=\frac{2^6.3^6+2^3.3^6+3^6}{-73}\)
\(D=\frac{3^6\left(2^6+2^3+1\right)}{-73}\)
\(D=\frac{3^6.73}{\left(-1\right).73}\)
\(D=-3^6=-729\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đổi hết thành ps rồi giải
gttđ > 0
cái này bn tự lm đc mà
cố lên
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)
\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)
~ Hok tốt ~
\(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^3.\left(-1\right)^5}{\frac{2}{5}.\left(-\frac{5}{12}\right)^2}\)
\(=\frac{\left(\frac{2}{3}\right)^3.\left(\frac{3}{4}\right)^3}{\frac{2}{5}.\left(\frac{5}{12}\right)^2}\)
\(=\frac{\left(\frac{2}{3}.\frac{3}{4}\right)^3}{\frac{2}{5}.\frac{5^2}{12^2}}\)
\(=\frac{\left(\frac{1}{2}\right)^3}{\frac{5}{72}}=\frac{1}{8}.\frac{72}{5}=\frac{9}{5}\)