\(F= \frac{5}{9} + \frac{4}{9} . \frac{3}{7} + \frac{4}{9}. \frac{4}{7}\)
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17 tháng 9 2020
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
9 tháng 2 2017
1. \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
= \(\frac{5}{9}\) .(\(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\) )
= \(\frac{5}{9}\) . 1 = \(\frac{5}{9}\)
25 tháng 4 2017
\(-\frac{3}{5}.\frac{5}{7}+-\frac{3}{5}.\frac{3}{7}+-\frac{3}{5}.\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}.2=-\frac{6}{5}\)
\(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{3}=\frac{1}{3}.2-\frac{4}{3}=\frac{2}{3}-\frac{4}{3}=-\frac{2}{3}\)
\(\frac{4}{19}.\frac{-3}{7}+-\frac{3}{7}.\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)
\(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\)
25 tháng 4 2017
\(a,6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}=6\frac{4}{5}-3\frac{4}{5}-1\frac{2}{3}=3-1\frac{2}{3}=\frac{4}{3}\)
\(b,6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)=6\frac{5}{7}-2\frac{5}{7}-1\frac{3}{4}=\frac{9}{4}\)
\(c,7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)=7\frac{5}{9}-3\frac{5}{9}-2\frac{3}{4}=4-2\frac{3}{4}=\frac{5}{4}\)
mk nghĩ là phần d như thế này cơ \(7\frac{5}{11}\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)=7\frac{5}{11}-3\frac{5}{11}-2\frac{3}{7}=4-2\frac{3}{7}=\frac{11}{7}\)
\(F=\frac{5}{9}+\frac{4}{9}\cdot\frac{3}{7}+\frac{4}{9}\cdot\frac{4}{7}\)
\(F=\frac{5}{9}+\frac{4}{9}\cdot\left(\frac{3}{7}+\frac{4}{7}\right)\)
\(F=\frac{5}{9}+\frac{4}{9}\cdot1\)
'\(F=\frac{5}{9}+\frac{4}{9}\)
\(F=1\)