\(\left[3,5+10\times\left(0,4\right)^2\right]\div\left[\left(0,5\right)^2-\left(\frac{1}{5}\right)^3+2,758\right]\)
Những câu hỏi liên quan
\(\left(2\frac{1}{3}+3,5\right)\div\left(-4\frac{1}{6}+3\frac{1}{7}\right)+75\%\)\(\left(-6\frac{3}{5}\div6-8\times12,5\%-\frac{2}{15}\div0,3\right)\times\left(-0,5\right)+1\)\(\left(2\frac{1}{3}+3\frac{1}{2}\right)\div\left(-4\frac{1}{6}+3\frac{1}{7}\right)+2\frac{15}{43}\)
25 tháng 8 2020 lúc 9:38
\(4\times\left(\frac{1}{4}\right)^2+25\times\left[\left(\frac{3}{4}\right)^3\div\left(\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)
\(2^3+3\times\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2\div\frac{1}{2}\right]-8\)
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
Tính:
a)\(\left\{\left[\left(6,2:0,31-\frac{5}{6}.0,9\right).0,2+0,15\right]:0,2\right\}:\left[\left(2+1\frac{4}{11}:0,1\right).\frac{1}{33}\right]\)
b)\(0,4\left(3\right)+0,6\left(2\right)-2\frac{1}{2}.\left[\left(\frac{1}{2}+\frac{1}{3}:0,5\left(8\right)\right)\right]:\frac{50}{53}\)
c)\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
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Tính nhanh :
(5+10+15+....+1000)+\(\left(\frac{2}{5}:0,5+2.\left(-0,4\right)\right):\left(\frac{1}{5}+\frac{1}{10}+....+\frac{1}{2012}\right)\)
\(2\frac{2}{3}\div\left\{\left[\left(3,72-0,02\times x\right)\frac{10}{37}\right]\div\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(2\frac{2}{3}:\left\{\left[\left(3,75-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=\frac{2}{3}\)
\(\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=4\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}=\frac{6}{5}\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]=1\)
\(\left(3,72-0,02.x\right)=\frac{37}{10}\)
\(0,02.x=0,02\)
\(x=1\)
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\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(\Rightarrow\frac{8}{3}:\left\{\left[\left(\frac{93}{25}-\frac{1}{50}.x\right)\frac{10}{37}\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{1}{5}\)
\(\Rightarrow\left\{\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{8}{3}:\frac{1}{5}=\frac{40}{3}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}=\frac{40}{3}+\frac{7}{15}=\frac{69}{5}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}=\frac{69}{5}-\frac{14}{5}=11\)
\(\Rightarrow\frac{93}{25}-\frac{1}{50}.x=11.\frac{5}{6}=\frac{55}{6}\)
\(\Rightarrow\frac{1}{50}.x=\frac{93}{25}-\frac{55}{6}=\frac{-817}{150}\)
\(\Rightarrow x=\frac{-817}{150}:\frac{1}{50}=\frac{-817}{3}\)
Ủng hộ tớ nha m.n?
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Tính:
\(\left(-2\frac{1}{5}\right)\times\left(\frac{-9}{11}\right)\times\left(-1\frac{1}{14}\right)\times\frac{2}{5}+\left(\frac{-4}{7}+\frac{6}{11}\right)\div\)\(\frac{-3}{5}\)
\(=\frac{11}{-5}\cdot\frac{-9}{11}\cdot\frac{15}{-14}\cdot\frac{2}{5}+-\frac{2}{77}\cdot\frac{5}{-3}\)
\(=\frac{9}{5}\cdot-\frac{15}{14}\cdot\frac{2}{5}+\frac{10}{231}\)
\(=-\frac{841}{1155}\)
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Tính:
\({\left( {\frac{{ - 2}}{3}} \right)^3};{\left( {\frac{{ - 3}}{5}} \right)^2};{\left( { - 0,5} \right)^3}; {\left( { - 0,5} \right)^2};\,{\left( {37,57} \right)^0};\,{\left( {3,57} \right)^1}\).
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^3} = \frac{{{{\left( { - 2} \right)}^3}}}{{{3^3}}} = \frac{{ - 8}}{{27}};\\{\left( {\frac{{ - 3}}{5}} \right)^2} = \frac{{{{\left( { - 3} \right)}^2}}}{{{5^2}}} = \frac{9}{{25}};\\{\left( { - 0,5} \right)^3} = {\left( {\frac{{ - 1}}{2}} \right)^3} = \frac{{{{\left( { - 1} \right)}^3}}}{{{2^3}}} = \frac{{ - 1}}{8};\\{\left( { - 0,5} \right)^2}=\frac{{{{\left( { - 1} \right)}^2}}}{{{2^2}}} = \frac{{1}}{4};\\\,{\left( {37,57} \right)^0} = 1;\,\\{\left( {3,57} \right)^1} = 3,57.\end{array}\)
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\frac{\left(\frac{2}{5}\right)^7\times 5^{7\:}+\left(\frac{9}{4}\right)^3\div \left(\frac{3}{16}\right)^3}{2^7\times 5^2+512}
Thay số thích hợp thay vào dấu “?” trong các câu sau:
a)\({\left[ {{{\left( {\frac{{ - 2}}{3}} \right)}^2}} \right]^5} = {\left( {\frac{{ - 2}}{3}} \right)^?};\) b)\({\left[ {{{\left( {0,4} \right)}^3}} \right]^3} = {\left( {0,4} \right)^?}\) c)\({\left[ {{{\left( {7,31} \right)}^3}} \right]^0} = ?\)
a)\({\left[ {{{\left( {\frac{{ - 2}}{3}} \right)}^2}} \right]^5} = {\left( {\frac{{ - 2}}{3}} \right)^{2.5}} = {\left( {\frac{{ - 2}}{3}} \right)^{10}}\)
Vậy dấu “?” bằng 10.
b) \({\left[ {{{\left( {0,4} \right)}^3}} \right]^3} = {\left( {0,4} \right)^{3.3}} = {\left( {0,4} \right)^9}\)
Vậy dấu “?” bằng 9.
c) \({\left[ {{{\left( {7,31} \right)}^3}} \right]^0} = 1\)
Vậy dấu “?” bằng 1.
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