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B=5/9.2,(37)+0,(5).3,(62)-2 căn bậc hai (-2/3)^2
B=5/9.235/99+5/9.359/99-4/3
B=5/9.(235/99+359/99)-4/3
B=5/9.6-4/3
B=10/3-4/3
B=2
K MK NHA
\(\left(-\frac{5}{12}\right):\frac{7}{3}-\left(-\frac{5}{12}\right):\frac{7}{4}=\left(-\frac{5}{12}\right):\left(\frac{7}{3}-\frac{7}{4}\right)=\left(-\frac{5}{12}\right):\frac{7}{12}=-\frac{5}{7}\)
\(\left[\left(\frac{2}{5}\right)^0\right].\frac{19}{13}-\left(\frac{7}{3}\right)^{2019}.\frac{3}{7}^{2019}\)
\(=\left(\frac{2}{5}\right)^0.\frac{19}{13}-\left(\frac{7}{3}.\frac{3}{7}\right)^{2019}\)
\(=1.\frac{19}{13}-1^{2019}\)
\(=1.\frac{19}{13}-1\)
\(=\frac{19}{13}-1\)
\(=\frac{6}{13}\)
Bài giải
a, \(\left(-\frac{5}{12}\right)\text{ : }\frac{7}{3}-\left(-\frac{5}{12}\right)\text{ : }\frac{7}{4}\)
\(=\left(-\frac{5}{12}\right)\text{ : }\frac{7}{3}-\left(-\frac{5}{12}\right)\text{ : }\frac{7}{4}\)
\(=\left(-\frac{5}{12}\right)\cdot\frac{3}{7}-\left(-\frac{5}{12}\right)\cdot\frac{4}{7}\)
\(=\frac{-15}{84}+\frac{20}{84}=\frac{5}{84}\)
b, \(\left[\left(\frac{2}{5}\right)^0\right]^{2020}\cdot\frac{19}{37}-\left(\frac{7}{3}\right)^{2019}\cdot\frac{3^{2019}}{7}\)
\(=1^{2020}\cdot\frac{19}{37}-\frac{7^{2019}}{3^{2019}}\cdot\frac{3^{2019}}{7}\)
\(=\frac{19}{37}-7^{2018}\)
\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)
\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)
\(=\left(\frac{2}{5}\right)^2\)
\(=\frac{4}{25}\)
\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)
\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)
\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)
\(=\frac{-7}{5}.\left(-10\right)\)
\(=14\)
a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)
\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)
\(=10:\left(\frac{-2}{3}\right)\)
\(=-15\)
b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\frac{-8}{27}+\frac{1}{10}\)
\(=\frac{-8}{3}+\frac{1}{10}\)
\(=\frac{-77}{30}\)
c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)
\(=\frac{-1}{3}\)
\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)
\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)
\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)
\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)
\(=-15\)
\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)
\(=-\frac{8}{3}+\frac{1}{10}\)
\(=-\frac{77}{30}\)
\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)
\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)
\(=\frac{1}{3}\)
Biến đổi :
\(\frac{125\cdot8^2}{30^6\cdot\left(-15\right)^2}=\frac{5^3\cdot\left(2^3\right)^2}{2^6\cdot3^6\cdot5^6\cdot3^2\cdot5^2}\)
\(=\frac{5^3\cdot2^6}{2^6\cdot3^8\cdot5^8}=\frac{1}{3^8\cdot5^5}\)
\(\Rightarrow\left(\frac{-2}{5}\right)^5:\left(\frac{125\cdot8^2}{30^6\cdot\left(-15\right)^2}\right)^2\)
\(=\frac{\left(-2\right)^5}{5^5}:\frac{1^2}{\left(3^8\cdot5^5\right)^2}\)
\(=\frac{\left(-2\right)^5\cdot3^{16}\cdot5^{10}}{5^5}\)
\(=\left(-2\right)^5\cdot3^{16}\cdot5^5\)
1.a) \(\left(31\frac{6}{13}+5\frac{9}{41}\right)-36\frac{6}{13}=\left(31+\frac{6}{13}+5+\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)\)
\(=\left(36+\frac{6}{13}-\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)=\left(36+\frac{6}{13}\right)-\left(36+\frac{6}{13}\right)-\frac{9}{41}=-\frac{9}{41}\)
b) \(\frac{5}{3}+\left(-\frac{2}{7}\right)-\left(-1,2\right)-\left|1.4-0,2\right|\)
\(=\frac{5}{3}-\frac{2}{7}+1,2-1,2=\frac{29}{21}\)
c) \(0,25+\frac{3}{5}-\left(\frac{1}{8}-\frac{2}{5}+1\frac{1}{4}\right)+\left|\frac{3}{5}\right|\)
\(=\frac{1}{4}+\frac{3}{5}-\frac{1}{8}+\frac{2}{5}-1-\frac{1}{4}+\frac{3}{5}\)
\(=\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}-1\right)+\frac{3}{5}-\frac{1}{8}=\frac{19}{40}\)
2) \(-\frac{3}{5}-x=0,75\)
=> \(-\frac{3}{5}-x=\frac{3}{4}\)
=> \(x=-\frac{3}{5}-\frac{3}{4}=\frac{-27}{20}\)
b) \(x+\frac{1}{3}=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
=> \(x+\frac{1}{3}=\frac{2}{5}+\frac{1}{3}\)
=> \(x=\frac{2}{5}\)
c) |2x - 4| + 1 = 5
=> |2x - 4| = 4
<=> \(\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
Giúp mình với nha cả nhả :<
Cả nhà làm vài ý thui cx được ạ :<
\(\left(\frac{-31}{37}\right)^5.\left(\frac{37}{62}\right)^5\)
\(=\left(\frac{-31}{37}.\frac{37}{62}\right)^5\)
\(=\left(\frac{-1}{2}\right)^5\)
\(=\frac{-1}{32}\)
\(\left(\frac{-31}{37}\right)^5.\left(\frac{37}{62}\right)^5\)
\(=\left(\frac{-31}{37}.\frac{37}{62}\right)^5\)
\(=\left(\frac{-1}{2}\right)^5\)
\(=\frac{-1}{32}\)
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