a)1/2x + 1/5 = 2/3x – 1/4
b) -1/5x – 2/3 = 1/10x + 5/6
c) x – 5/4 = 1/3 – 3/4x
d) 3/2x – 1 1/2 = x – 3/4
e) -4/5 + 2x = /3 – 2/3x
a)1/2x + 1/5 = 2/3x – 1/4
b) -1/5x – 2/3 = 1/10x + 5/6
c) x – 5/4 = 1/3 – 3/4x
d) 3/2x – 1 1/2 = x – 3/4
e) -4/5 + 2x = /3 – 2/3x
Đáp án:
$\begin{array}{l}
a)\dfrac{1}{2}x + \dfrac{1}{5} = \dfrac{2}{3}x – \dfrac{1}{4}\\
\Leftrightarrow \dfrac{2}{3}x – \dfrac{1}{2}x = \dfrac{1}{5} + \dfrac{1}{4}\\
\Leftrightarrow \left( {\dfrac{2}{3} – \dfrac{1}{2}} \right).x = \dfrac{9}{{20}}\\
\Leftrightarrow \dfrac{1}{6}.x = \dfrac{9}{{20}}\\
\Leftrightarrow x = \dfrac{9}{{20}}.6\\
\Leftrightarrow x = \dfrac{{27}}{{10}}\\
Vậy\,x = \dfrac{{27}}{{10}}\\
b) – \dfrac{1}{5}x – \dfrac{2}{3} = \dfrac{1}{{10}}x + \dfrac{5}{6}\\
\Leftrightarrow \dfrac{1}{{10}}x + \dfrac{1}{5}.x = – \dfrac{2}{3} – \dfrac{5}{6}\\
\Leftrightarrow \left( {\dfrac{1}{{10}} + \dfrac{1}{5}} \right).x = \dfrac{{ – 4 – 5}}{6}\\
\Leftrightarrow \dfrac{3}{{10}}.x = \dfrac{{ – 9}}{6}\\
\Leftrightarrow x = \dfrac{{ – 9}}{6}.\dfrac{{10}}{3}\\
\Leftrightarrow x = \dfrac{{ – 5}}{3}\\
Vậy\,x = – \dfrac{5}{3}\\
c)x – \dfrac{5}{4} = \dfrac{1}{3} – \dfrac{3}{4}.x\\
\Leftrightarrow x + \dfrac{3}{4}.x = \dfrac{1}{3} + \dfrac{5}{4}\\
\Leftrightarrow \left( {1 + \dfrac{3}{4}} \right).x = \dfrac{{4 + 15}}{{12}}\\
\Leftrightarrow \dfrac{7}{4}.x = \dfrac{{19}}{{12}}\\
\Leftrightarrow x = \dfrac{{19}}{{12}}.\dfrac{4}{7}\\
\Leftrightarrow x = \dfrac{{19}}{{21}}\\
Vậy\,x = \dfrac{{19}}{{21}}\\
d)\dfrac{3}{2}.x – 1\dfrac{1}{2} = x – \dfrac{3}{4}\\
\Leftrightarrow \dfrac{3}{2}.x – x = 1\dfrac{1}{2} – \dfrac{3}{4}\\
\Leftrightarrow \left( {\dfrac{3}{2} – 1} \right).x = \dfrac{3}{2} – \dfrac{3}{4}\\
\Leftrightarrow \dfrac{1}{2}.x = \dfrac{3}{4}\\
\Leftrightarrow x = \dfrac{3}{4}.2\\
\Leftrightarrow x = \dfrac{3}{2}\\
Vậy\,x = \dfrac{3}{2}\\
e) – \dfrac{4}{5} + 2x = \dfrac{1}{3} – \dfrac{2}{3}.x\\
\Leftrightarrow 2x + \dfrac{2}{3}.x = \dfrac{1}{3} + \dfrac{4}{5}\\
\Leftrightarrow \left( {2 + \dfrac{2}{3}} \right).x = \dfrac{{17}}{{15}}\\
\Leftrightarrow \dfrac{8}{3}.x = \dfrac{{17}}{{15}}\\
\Leftrightarrow x = \dfrac{{17}}{{15}}.\dfrac{3}{8}\\
\Leftrightarrow x = \dfrac{{17}}{{40}}\\
Vậy\,x = \dfrac{{17}}{{40}}
\end{array}$