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Từ đề bài ta có:
4/3x - 1/3 = (2x-1) : 3/5
=> 4/3x -1/3 = (2x-1) * 5/3
=> 4/3x -1/3= 10/3x - 5/3
Chuyển vế đổi dấu
Ta được:
=> -2x = -4/3
=> x= 2/3
Vậy x= 2/3
\(1\frac{1}{3}x=\left(2x-1\right):\left(1-\frac{2}{5}\right)\)
\(\frac{4}{3}x=\left(2x-1\right):\left(\frac{3}{5}\right)\)
\(\frac{4}{3}x.\frac{3}{5}=\left(2x-1\right)\)
\(\frac{4}{5}x=\left(2x-1\right)\)
\(x=\left(2x-1\right):\frac{4}{5}\)
\(x=\left(2x-1\right).\frac{5}{4}\)
\(x=2x.\frac{5}{4}-1.\frac{5}{4}\)
\(x=\)\(2x.\frac{5}{4}-\frac{5}{4}\)
\(x=2.\frac{5}{4}.x-\frac{5}{4}\)
\(x=\left(\frac{5}{2}.x\right)-\frac{5}{4}\)
\(x=\frac{5}{2}-\frac{5}{4}.x-\frac{5}{4}\)
\(x=\frac{5}{4}.x-\frac{5}{4}\)
\(x=x\left(\frac{5}{4}-\frac{5}{4}\right)\)
\(x=x.0\)
\(=>x=0\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
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)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
\(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(x+3\frac{1}{7}\right)+7\frac{1}{2}=1\frac{69}{86}\)
\(\left(\frac{7}{3}+\frac{7}{2}\right):\left(x+\frac{22}{7}\right)+\frac{15}{2}=\frac{155}{86}\)
\(\left(\frac{14}{6}+\frac{21}{6}\right):\left(x+\frac{22}{7}\right)+\frac{15}{2}=\frac{155}{86}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{155}{86}-\frac{15}{2}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{155}{86}-\frac{645}{86}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{-245}{43}\)
\(x+\frac{22}{7}=\frac{35}{6}:\frac{-245}{43}=\frac{35}{6}\cdot\frac{-43}{245}\)
\(x+\frac{22}{7}=\frac{-43}{42}\)
\(x=\frac{-43}{42}-\frac{22}{7}=\frac{-43}{42}-\frac{132}{42}\)
\(x=\frac{-25}{6}\)