a,\((\frac{32}{81})^2.\left(\frac{-9}{8}\right)^5.\left(-4\right)^3̣\)
b,\(\left(-25\right)^5:125^2:\left(-5\right)^3\)
c,\(\left(\frac{8}{27}\right)^3:\left(\frac{-2}{3}\right)^8\)
a, \(\left(7+\frac{7}{5}-\frac{2}{3}\right)-\left(4+\frac{4}{5}+\frac{3}{8}\right)+\left(3-\frac{3}{5}+\frac{2}{3}+\frac{3}{8}\right)\)
b.\(-\frac{13}{25}.\frac{5}{32}.\left(\frac{25}{-13}\right).\left(-64\right)\)
c.\(\frac{7}{80}:\left(-1\frac{3}{4}\right)-\frac{2}{9}:\left(8-\frac{8}{3}\right)-\left(-\frac{5}{24}\right)\left(-\frac{10}{3}+2\frac{8}{15}\right)\)
Tính
a) \(\left( {\frac{4}{5} - 1} \right):\frac{3}{5} - \frac{2}{3}.0,5\)
b) \(1 - {\left( {\frac{5}{9} - \frac{2}{3}} \right)^2}:\frac{4}{{27}}\)
c)\(\left[ {\left( {\frac{3}{8} - \frac{5}{{12}}} \right).6 + \frac{1}{3}} \right].4\)
d) \(0,8:\left\{ {0,2 - 7.\left[ {\frac{1}{6} + \left( {\frac{5}{{21}} - \frac{5}{{14}}} \right)} \right]} \right\}\)
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)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
a)\(\left(1-\frac{2}{5}\right)^2+\left|\frac{-3}{5}\right|+\frac{-7}{10}\)
b)\(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}\)
c)\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
Rút gọn
a)\(\frac{25^{5^{ }}.5^{10}}{100^5}.\frac{4^6.9^5+6^9.120}{8^4.3^{12^{ }}-6^{11}}\)
b)\(\left(\frac{4}{9}+\frac{1}{3}\right)^2+\left(\frac{3}{4}\right)^2:\left(\frac{3}{4}\right)^2:\left(-\frac{2}{3}\right)^3\)
c) (273 : 33) : \(\left[\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5\right]\)
CÁC BẠN GIÚP MIH VỚI
Tính giá trị biểu thức
\(1.A=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{1}{7}+\left[\frac{-14}{30}\right]\)
\(2.B=\left(\frac{5}{8}-\frac{4}{12}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left[\frac{-3}{2}\right]+\frac{7}{-15}\)
\(3.C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(4.D=\frac{1}{9}-\left[\frac{-5}{23}\right]-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{14}-\frac{7}{30}\)
\(5.E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)+\left(\frac{12}{17}+\frac{5}{18}+\frac{7}{5}\right)\)
\(6.F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{12}{17}-\frac{15}{14}+\frac{1}{4}\right)\)
\(7.G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(8.H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{29}{27}\right)-\frac{2}{27}\)
\(9.K=\frac{1}{16}-\frac{5}{21}+\left(\frac{-1}{16}+\frac{-3}{5}-\frac{-5}{21}\right)+\frac{-2}{5}+\frac{3}{4}\)
\(10.L=\frac{7}{12}+\frac{15}{14}-\left(\frac{14}{22}+\frac{-1}{14}+\frac{5}{21}\right)-\frac{-5}{21}+\frac{3}{5}\)
yutyugubhujyikiu
\(A=\frac{2^{12}x3^5-4^6x9^2}{\left(2^2x3\right)^6+8^4x3^5}-\frac{5^{10}x7^3-25^5x49^2}{\left(125x7\right)^3+5^9x14^3}\)
\(B=\frac{\left(\frac{-1}{2}\right)^3-\left(\frac{3}{4}\right)^3x\left(-2\right)^2}{2x\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}}\)
\(C=2^2+3\left(\frac{1}{2}\right)^0-2^{-2}+\left[\left(-2^2\right):\frac{1}{2}\right]:8\)
A= \(\frac{\left(81,624:\frac{16}{3}-4,505\right)^2+125.\frac{3}{4}}{\left\{\left[\left(\frac{11}{25}\right)^2:0,88+3,53\right]^2-\left(2,75\right)^2\right\}:\frac{13}{25}}\) B= \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
Thực hiện các phép tính sau
\(A=\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)|
\(B=-3-\frac{2}{3}+\frac{3}{5}\left(-\frac{10}{9}-\frac{25}{3}\right)-\frac{5}{6}\)
\(C=\left(\frac{12}{35}-\frac{6}{7}+\frac{18}{14}\right):\frac{6}{-7}-\frac{-2}{5}-1\)
\(D=\left[\frac{-54}{64}-\left(\frac{1}{9}:\frac{8}{27}\right):\frac{-1}{3}\right]:\frac{-81}{128}\)
\(E=\left[\frac{193}{-17}\left(\frac{2}{193}-\frac{3}{386}\right)+\frac{11}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right)\frac{1931}{25}+\frac{9}{2}\right]\)
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A = \(\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)
A = \(\left(-\frac{1}{4}\right)+\frac{5}{6}-\frac{7}{12}\)
A = \(\frac{7}{12}-\frac{7}{12}\)
A = \(0\).
Mình làm câu A thôi nhé.
Chúc bạn học tốt!
Tính giá trị các biểu thức.
a)\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}};\)
b)\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}};\)
c)\(\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}};\)
d)\(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}}.\)
a)
\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}} = \frac{{{{\left( {{2^2}} \right)}^3}.{{\left( {{3^2}} \right)}^7}}}{{{{\left( {{3^3}} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} =\frac{2^{2.3}.3^{2.7}}{3^{3.5}.2^{2.3}}= \frac{{{2^6}{{.3}^{14}}}}{{{3^{15}}{{.2}^6}}} = \frac{1}{3}\)
b)
\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}} =\frac{(-2)^{3+7}}{3.(2^2)^6}= \frac{{{{\left( { - 2} \right)}^{10}}}}{{3.{{\left( {{2^{2.6}}} \right)}}}} = \frac{{{2^{10}}}}{{{{3.2}^{12}}}} = \frac{1}{{{{3.2}^2}}} = \frac{1}{{12}}\)
c)
\(\begin{array}{l}\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left[ {{{\left( {0,3} \right)}^2}} \right]}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,3} \right)}^6}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}}\\ = \frac{{{{\left( {0,3} \right)}^2}}}{{{{\left( {0,2} \right)}^2}}} = \frac{{0,9}}{{0,4}} = \frac{9}{4}\end{array}\)
d)
Cách 1: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{8 + 16 + 32}}{{49}} = \frac{{56}}{{49}} = \frac{8}{7}\)
Cách 2: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)