Bài 6:
\(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Rightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=-\dfrac{15}{28}+\dfrac{11}{13}\)
\(\Rightarrow x=-\dfrac{15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}=-\dfrac{5}{12}\)
Bài 2:
a) \(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{5}\)
b) \(\Rightarrow x=\dfrac{3}{7}-\dfrac{1}{4}-\dfrac{3}{5}=-\dfrac{59}{140}\)
Bài 6:
\(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
<=> \(\dfrac{11}{13}-\dfrac{5}{42}+x=-\dfrac{15}{28}+\dfrac{11}{13}\)
<=> \(x=-\dfrac{15}{28}+\dfrac{11}{13}-\dfrac{11}{13}-\dfrac{5}{42}\)
<=> \(x=-\dfrac{15}{28}-\dfrac{5}{42}\)
<=> \(x=\dfrac{-55}{84}\)
Bài 7:
a. \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
<=> \(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
<=> \(x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}\)
<=> \(x=\dfrac{2}{5}\)
b. \(\dfrac{3}{7}-x=\dfrac{1}{4}-\left(-\dfrac{3}{5}\right)\)
<=> \(\dfrac{3}{7}-x=\dfrac{1}{4}+\dfrac{3}{5}\)
<=> \(-x=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{3}{7}\)
<=> \(-x=\dfrac{59}{140}\)
<=> \(x=\dfrac{-59}{140}\)
Bài 8:
a: Ta có: \(x+\dfrac{3}{10}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}-\dfrac{3}{10}=\dfrac{-4}{30}-\dfrac{9}{30}=\dfrac{-13}{30}\)
b: Ta có: \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(\Leftrightarrow x+\dfrac{5}{6}=\dfrac{2}{3}+\dfrac{2}{5}=\dfrac{16}{15}\)
\(\Leftrightarrow x=\dfrac{16}{15}-\dfrac{5}{6}=\dfrac{96-75}{90}=\dfrac{7}{30}\)
