\(2\frac{2}{3}\div\left\{\left[\left(3,72-0,02\times x\right)\frac{10}{37}\right]\div\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
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KN
25 tháng 8 2020
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
DH
27 tháng 7 2020
\(\left[\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8+1,75\right]:0,05=235\)
\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8+1,75=11,75\)
\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8=10\)
\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8=10\)
\(\Leftrightarrow6\frac{3}{7}-\frac{0,75x-2}{0,35}=\frac{25}{7}\)
\(\Leftrightarrow\frac{0,75x-2}{0,35}=\frac{20}{7}\)
\(\Leftrightarrow0,75x-2=1\)
\(\Leftrightarrow0,75x=3\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
EG
17 tháng 8 2018
\(=\frac{11}{-5}\cdot\frac{-9}{11}\cdot\frac{15}{-14}\cdot\frac{2}{5}+-\frac{2}{77}\cdot\frac{5}{-3}\)
\(=\frac{9}{5}\cdot-\frac{15}{14}\cdot\frac{2}{5}+\frac{10}{231}\)
\(=-\frac{841}{1155}\)
11 tháng 7 2019
1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(2\frac{2}{3}:\left\{\left[\left(3,75-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=\frac{2}{3}\)
\(\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=4\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}=\frac{6}{5}\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]=1\)
\(\left(3,72-0,02.x\right)=\frac{37}{10}\)
\(0,02.x=0,02\)
\(x=1\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(\Rightarrow\frac{8}{3}:\left\{\left[\left(\frac{93}{25}-\frac{1}{50}.x\right)\frac{10}{37}\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{1}{5}\)
\(\Rightarrow\left\{\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{8}{3}:\frac{1}{5}=\frac{40}{3}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}=\frac{40}{3}+\frac{7}{15}=\frac{69}{5}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}=\frac{69}{5}-\frac{14}{5}=11\)
\(\Rightarrow\frac{93}{25}-\frac{1}{50}.x=11.\frac{5}{6}=\frac{55}{6}\)
\(\Rightarrow\frac{1}{50}.x=\frac{93}{25}-\frac{55}{6}=\frac{-817}{150}\)
\(\Rightarrow x=\frac{-817}{150}:\frac{1}{50}=\frac{-817}{3}\)
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