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a) \(-\frac{48}{625}\)
b) không có giá trị
c) 1:125
d) \(-\frac{1053}{500}\)
e) 1
g) -13
a) \(-\dfrac{48}{625}\)
b) \(\varnothing\)
c) \(\dfrac{1}{125}\)
d) \(-\dfrac{1053}{500}\)
e) 1
f) -13
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(a,\left(\frac{1}{\sqrt{625}}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{\sqrt{25}}-1\right)\)
\(=\left(\frac{1}{25}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{5}-1\right)\)
\(=\frac{31}{25}:\left(-\frac{29}{25}\right)\)
\(=\frac{31}{25}.\frac{-25}{29}\)
\(=-\frac{31}{29}\)
\(b,\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)\right]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)
\(=\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}.\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}.\frac{19}{4}\right)\)
\(=\left(\frac{109}{6}-\frac{13}{10}\right):\frac{19}{3}\)
\(=\frac{253}{15}.\frac{3}{19}\)
\(=\frac{253}{95}\)
Số to :v
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)
\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)
=>x=0
c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)
d: \(\Leftrightarrow x^8=x^7\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)
hay x=1
f: =>x-1=20
hay x=21