Giá trị của \(x\) thỏa mãn \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\) là
\(\dfrac{6}{11}.\)\(\dfrac{7}{11}.\)\(\dfrac{8}{11}.\)\(\dfrac{9}{11}.\)Hướng dẫn giải:\(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(\left(\dfrac{1}{3}+\dfrac{2}{5}\right)x=\dfrac{2}{5}\)
\(\dfrac{11}{15}x=\dfrac{2}{5}\)
\(x=\dfrac{2}{5}:\dfrac{11}{15}\)
\(x=\dfrac{6}{11}.\)
