\(\frac{11}{36}-\left(-\frac{7}{24}\right)\)
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\(=3-1+\frac{1}{2^2.2}\)
\(=2+\frac{1}{8}=\frac{17}{8}\)
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{2^6}\)\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^{21-12}\)\(=\left(\frac{3}{7}\right)^9\)
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^9.\left(\frac{3}{7}\right)^{12}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^9.\left(\frac{3}{7}\right)^{12}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^9=\frac{3^9}{7^9}=\frac{19683}{40353607}\)
7.(\(\frac{3}{8}+\frac{11}{7}\)) - 11.\(\frac{7}{17}\) + 7.\(\frac{5}{8}\)
= 7(\(\frac{3}{8}\) +\(\frac{11}{7}\) -\(\frac{11}{7}+\frac{5}{8}\))
=7(\(\frac{3}{8}+\frac{5}{8}\))
= 7.1
=7
\(\frac{11}{36}-\left(-\frac{7}{24}\right)\)
\(=\frac{11}{36}+\frac{7}{24}\)
\(=\frac{22}{72}+\frac{21}{72}\)
\(=\frac{43}{72}\)
= 43/72 nha