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\(=\frac{\left(3.7\right)^4}{3^3.\left(-7\right)^3}+7\)
\(=\frac{3^4.7^4}{3^3.\left(-7\right)^3}+7\)
=\(\) 3.(-7)+7
= -14
a) \(1\frac{3}{19}+\frac{8}{21}-\frac{3}{19}+0.5+\frac{13}{21}\)
\(=\left(1\frac{3}{19}-\frac{3}{19}\right)+\left(\frac{8}{21}+\frac{13}{21}\right)+0.5\)
\(=1+1+0.5=2.5\)
b) \(\left(-\frac{3}{4}+\frac{2}{7}\right):\frac{3}{7}+\left(\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=0:\frac{3}{7}=0\)
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
\(a.\)
\(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}\)
\(=\left(\frac{15}{33}+\frac{18}{33}\right)+\left(\frac{13}{20}+\frac{7}{20}\right)\)
\(=\frac{33}{33}+\frac{20}{20}\)
\(=1+1=2\)
\(b.\)
\(2\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{21}\right)\)
\(=\frac{5}{2}+\frac{4}{7}:\left(-\frac{8}{21}\right)\)
\(=\frac{5}{2}+\frac{4}{7}.\left(-\frac{21}{8}\right)\)
\(=\frac{5}{2}+\frac{1}{1}.\left(-\frac{3}{2}\right)\)
\(=\frac{5}{2}-\frac{3}{2}\)
\(=1\)
\(c.\)
\(\left(-\frac{1}{2}\right)^3+\frac{1}{2}:5\)
\(=-\frac{1}{8}+\frac{1}{2}.\frac{1}{5}\)
\(=-\frac{1}{8}+\frac{1}{10}\)
\(=-\frac{1}{40}\)
a) \(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}=\left(\frac{15}{33}+\frac{18}{33}\right)+\left(\frac{7}{20}+\frac{13}{20}\right)\) = 1 + 1 = 2
b) \(2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)=\frac{5}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)=\frac{5}{2}+\frac{-3}{2}\) = 1
c) \(\left(\frac{-1}{2}\right)\)3 + \(\frac{1}{2}\) : 5 = \(\frac{-1}{8}+\frac{1}{10}\) = \(\frac{-1}{40}\)
Chúc bạn học tốt!
a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5=2,5\)
b)\(\frac{3}{7}\cdot19\frac{1}{3}-\frac{3}{7}\cdot33\frac{1}{3}\)
\(=\frac{3}{7}\left(19\frac{1}{3}-33\frac{1}{3}\right)=\frac{3}{7}\cdot\left(-14\right)=-6\)
c) \(9\left(-\frac{1}{3}\right)^3+\frac{1}{3}=9\cdot\left(\frac{-1}{27}\right)+\frac{1}{3}=-\frac{1}{3}+\frac{1}{3}=0\)
d) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)=-\frac{7}{5}:\left(15\frac{1}{4}-25\frac{1}{4}\right)=-\frac{7}{5}:\left(-10\right)=14\)
a) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(\frac{27}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b) \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\frac{58}{3}-\frac{3}{7}.\frac{100}{3}\)
\(=\frac{3}{7}.\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{7}.\left(-14\right)\)
\(=-6\)
c) \(9.\left(-\frac{1}{3}\right)^3+\frac{1}{3}\)
\(=9.\left(-\frac{1}{27}\right)+\frac{1}{3}\)
\(=-\frac{1}{3}+\frac{1}{3}\)
\(=0\)
d) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)
\(=\frac{61}{4}:\left(-\frac{5}{7}\right)-\frac{101}{4}:\left(-\frac{5}{7}\right)\)
\(=\left(\frac{61}{4}-\frac{101}{4}\right):\left(-\frac{5}{7}\right)\)
\(=\left(-10\right):\left(-\frac{5}{7}\right)\)
\(=14\)
^...^ 
^_^
\(\left(5-\frac{2}{3}+\frac{3}{7}\right):\left(24-25+\frac{4}{21}-\frac{8}{21}\right)\)
\(\left(\frac{5.21-14+9}{21}\right):\left(\frac{-21-4}{21}\right)\)=\(\left(\frac{5.21-5}{21}\right).\left(\frac{21}{-25}\right)\)=\(\frac{5\left(21-1\right)}{\left(-5\right).5}=\frac{20}{-5}\)
=-4
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) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
\(\frac{21^4}{27\cdot(-343)}+7\)
\(=\frac{(3\cdot7)^4}{3^3\cdot(-7)^3}+7\)
\(=\frac{3^4\cdot7^4}{3^3\cdot(-7)^3}+7\)
\(=3\cdot(-7)+7=-14\)