\(c)x+\dfrac{4}{5}=\dfrac{6}{20}-\dfrac{-7}{3}\)
\(x+\dfrac{4}{5}=\dfrac{3}{10}+\dfrac{7}{3}\)
\(x=\dfrac{3}{10}+\dfrac{7}{3}-\dfrac{4}{5}\)
\(x=\dfrac{9}{30}+\dfrac{70}{30}-\dfrac{24}{30}
\(x=\dfrac{11}{6}
Vậy \(x=\dfrac{11}{6}
Bài \(5:\)
\(a)\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)
\(⇒\dfrac{x-3}{13}+\dfrac{x-3}{14}-\dfrac{x-3}{15}-\dfrac{x-3}{16}=0\)
\(⇒(x-3).(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16})=0\)
Vì \(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16} \ne 0\) nên \(x-3=0\)
\(⇒x=3\)
Vậy \(x=3\)
\(b)\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(⇒(\dfrac{x+1}{65}+1)+(\dfrac{x+3}{63}+1)=(\dfrac{x+5}{61}+1)+(\dfrac{x+7}{59}+1)\)
\(⇒\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)
\(⇒\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)
\(⇒(x+66).(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59})=0\)
Vì \(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}≠0\) nên \(x+66=0\)
\(⇒x=-66\)
Vậy \(x=-66\)