Question

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Answer 1

$a, \frac{17}{6} - (x - \frac{7}{6}) = \frac{7}{4} = > \frac{17}{6} - x + \frac{7}{6} = \frac{7}{4} = > 4 - x = \frac{7}{4} = > x = 4 - \frac{7}{4} = > x = \frac{9}{4} V ậ y x = \frac{9}{4} b, \frac{3}{8} - (2 x - \frac{1}{2}) : {(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{3} = > (2 x - \frac{1}{2}) : {(\frac{1}{2})}^{2} = \frac{3}{8} - \frac{1}{8} = > (2 x - \frac{1}{2}) : \frac{1}{4} = \frac{1}{4} = > 2 x - \frac{1}{2} = \frac{1}{4} \times \frac{1}{4} = > 2 x - \frac{1}{2} = \frac{1}{16} = > 2 x = \frac{1}{16} + \frac{1}{2} = > 2 x = \frac{9}{16} = > x = \frac{9}{32} V ậ y x = \frac{9}{32}$

$c, {(2 x - \frac{1}{2})}^{2} = \frac{16}{25} = > 2 x - \frac{1}{2} = \frac{4}{5} h o ặ c 2 x - \frac{1}{2} = \frac{- 4}{5} = > 2 x = \frac{4}{5} + \frac{1}{2} h o ặ c 2 x = \frac{- 4}{5} + \frac{1}{2} = > 2 x = \frac{13}{10} h o ặ c 2 x = \frac{- 3}{10} = > x = \frac{13}{20} h o ặ c x = \frac{- 3}{20} v ậ y S = \{\frac{- 3}{20}; \frac{13}{20}\}$ .

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Answer 2

$a, \frac{17}{6} - (x - \frac{7}{6}) = \frac{7}{4} = > \frac{17}{6} - x + \frac{7}{6} = \frac{7}{4} = > 4 - x = \frac{7}{4} = > x = 4 - \frac{7}{4} = > x = \frac{9}{4} V ậ y x = \frac{9}{4} b, \frac{3}{8} - (2 x - \frac{1}{2}) : {(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{3} = > (2 x - \frac{1}{2}) : {(\frac{1}{2})}^{2} = \frac{3}{8} - \frac{1}{8} = > (2 x - \frac{1}{2}) : \frac{1}{4} = \frac{1}{4} = > 2 x - \frac{1}{2} = \frac{1}{4} \times \frac{1}{4} = > 2 x - \frac{1}{2} = \frac{1}{16} = > 2 x = \frac{1}{16} + \frac{1}{2} = > 2 x = \frac{9}{16} = > x = \frac{9}{32} V ậ y x = \frac{9}{32}$

$c, {(2 x - \frac{1}{2})}^{2} = \frac{16}{25} = > 2 x - \frac{1}{2} = \frac{4}{5} h o ặ c 2 x - \frac{1}{2} = \frac{- 4}{5} = > 2 x = \frac{4}{5} + \frac{1}{2} h o ặ c 2 x = \frac{- 4}{5} + \frac{1}{2} = > 2 x = \frac{13}{10} h o ặ c 2 x = \frac{- 3}{10} = > x = \frac{13}{20} h o ặ c x = \frac{- 3}{20} v ậ y S = \{\frac{- 3}{20}; \frac{13}{20}\}$ .

Answer 3

\[\begin{array}{l}
a)\frac{{17}}{6} - \left( {x - \frac{7}{6}} \right) = \frac{7}{4}\\
= > \frac{{17}}{6} - x + \frac{7}{6} = \frac{7}{4}\\
= > \frac{{24}}{6} - x = \frac{7}{4}\\
= > x = \frac{{24}}{6} - \frac{7}{4}\\
= > x = \frac{9}{4}\\
vay:x = \frac{9}{4}\\
{\rm{b)}}\frac{3}{8} - \left( {2x - \frac{1}{2}} \right):{\left( {\frac{1}{2}} \right)^2} = {\left( {\frac{1}{2}} \right)^3}\\
= > \frac{3}{8} - \left( {2x - \frac{1}{2}} \right):\frac{1}{4} = \frac{1}{8}\\
= > \frac{3}{8} - \left( {2x - \frac{1}{2}} \right).4 = \frac{1}{8}\\
= > \frac{3}{8} - \left( {8x - 2} \right) = \frac{1}{8}\\
= > 8x - 2 = \frac{3}{8} - \frac{1}{8}\\
= > 8x - 2 = \frac{1}{4}\\
= > 8x = 2 + \frac{1}{4}\\
= > 8x = \frac{9}{4}\\
= > x = \frac{9}{4}:8\\
= > x = \frac{9}{{32}}\\
vay:x = \frac{9}{{32}}\\
c){\left( {2x - \frac{1}{2}} \right)^2} = \frac{{16}}{{25}}\\
= > \left[ \begin{array}{l}
2x - \frac{1}{2} = \frac{4}{5}\\
2x - \frac{1}{2} = \frac{{ - 4}}{5}
\end{array} \right.\\
= > \left[ \begin{array}{l}
2x = \frac{{13}}{{10}}\\
2x = \frac{{ - 3}}{{10}}
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = \frac{{13}}{{20}}\\
x = \frac{{ - 3}}{{20}}
\end{array} \right.\\
vay:S = \{ \frac{{13}}{{20}};\frac{{ - 3}}{{20}}\}
\end{array}\]

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Answer 4

\[\begin{array}{l}
a)\frac{{17}}{6} - \left( {x - \frac{7}{6}} \right) = \frac{7}{4}\\
= > \frac{{17}}{6} - x + \frac{7}{6} = \frac{7}{4}\\
= > \frac{{24}}{6} - x = \frac{7}{4}\\
= > x = \frac{{24}}{6} - \frac{7}{4}\\
= > x = \frac{9}{4}\\
vay:x = \frac{9}{4}\\
{\rm{b)}}\frac{3}{8} - \left( {2x - \frac{1}{2}} \right):{\left( {\frac{1}{2}} \right)^2} = {\left( {\frac{1}{2}} \right)^3}\\
= > \frac{3}{8} - \left( {2x - \frac{1}{2}} \right):\frac{1}{4} = \frac{1}{8}\\
= > \frac{3}{8} - \left( {2x - \frac{1}{2}} \right).4 = \frac{1}{8}\\
= > \frac{3}{8} - \left( {8x - 2} \right) = \frac{1}{8}\\
= > 8x - 2 = \frac{3}{8} - \frac{1}{8}\\
= > 8x - 2 = \frac{1}{4}\\
= > 8x = 2 + \frac{1}{4}\\
= > 8x = \frac{9}{4}\\
= > x = \frac{9}{4}:8\\
= > x = \frac{9}{{32}}\\
vay:x = \frac{9}{{32}}\\
c){\left( {2x - \frac{1}{2}} \right)^2} = \frac{{16}}{{25}}\\
= > \left[ \begin{array}{l}
2x - \frac{1}{2} = \frac{4}{5}\\
2x - \frac{1}{2} = \frac{{ - 4}}{5}
\end{array} \right.\\
= > \left[ \begin{array}{l}
2x = \frac{{13}}{{10}}\\
2x = \frac{{ - 3}}{{10}}
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = \frac{{13}}{{20}}\\
x = \frac{{ - 3}}{{20}}
\end{array} \right.\\
vay:S = \{ \frac{{13}}{{20}};\frac{{ - 3}}{{20}}\}
\end{array}\]

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