Giải thích các bước giải: Ta có: \(\begin{array}{l}b,\\x + \dfrac{1}{3} = \dfrac{2}{5} – \left( { – \dfrac{1}{3}} \right)\\ \Leftrightarrow x + \dfrac{1}{3} = \dfrac{2}{5} + \dfrac{1}{3}\\ \Leftrightarrow x = \dfrac{2}{5}\\c,\\ – \dfrac{2}{3}x + \dfrac{5}{7} = \dfrac{3}{{10}}\\ \Leftrightarrow – \dfrac{2}{3}x = \dfrac{3}{{10}} – \dfrac{5}{7}\\ \Leftrightarrow – \dfrac{2}{3}x = – \dfrac{{29}}{{70}}\\ \Leftrightarrow x = \left( { – \dfrac{{29}}{{70}}} \right):\left( { – \dfrac{2}{3}} \right)\\ \Leftrightarrow x = \dfrac{{87}}{{140}}\\d,\\\dfrac{3}{7} – x = \dfrac{1}{4} – \left( { – \dfrac{3}{5}} \right)\\ \Leftrightarrow \dfrac{3}{7} – x = \dfrac{1}{4} + \dfrac{3}{5}\\ \Leftrightarrow \dfrac{3}{7} – x = \dfrac{{17}}{{20}}\\ \Leftrightarrow x = \dfrac{3}{7} – \dfrac{{17}}{{20}}\\ \Leftrightarrow x = – \dfrac{{59}}{{140}}\end{array}\) Em xem lại đề câu a nhé! Reply
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
x + \dfrac{1}{3} = \dfrac{2}{5} – \left( { – \dfrac{1}{3}} \right)\\
\Leftrightarrow x + \dfrac{1}{3} = \dfrac{2}{5} + \dfrac{1}{3}\\
\Leftrightarrow x = \dfrac{2}{5}\\
c,\\
– \dfrac{2}{3}x + \dfrac{5}{7} = \dfrac{3}{{10}}\\
\Leftrightarrow – \dfrac{2}{3}x = \dfrac{3}{{10}} – \dfrac{5}{7}\\
\Leftrightarrow – \dfrac{2}{3}x = – \dfrac{{29}}{{70}}\\
\Leftrightarrow x = \left( { – \dfrac{{29}}{{70}}} \right):\left( { – \dfrac{2}{3}} \right)\\
\Leftrightarrow x = \dfrac{{87}}{{140}}\\
d,\\
\dfrac{3}{7} – x = \dfrac{1}{4} – \left( { – \dfrac{3}{5}} \right)\\
\Leftrightarrow \dfrac{3}{7} – x = \dfrac{1}{4} + \dfrac{3}{5}\\
\Leftrightarrow \dfrac{3}{7} – x = \dfrac{{17}}{{20}}\\
\Leftrightarrow x = \dfrac{3}{7} – \dfrac{{17}}{{20}}\\
\Leftrightarrow x = – \dfrac{{59}}{{140}}
\end{array}\)
Em xem lại đề câu a nhé!