Question

Tìm x

\(\dfrac{1}{2}\)x + \(\dfrac{2}{3}\) x-1 = \(-3\dfrac{1}{3}\)

Answer 1

\(\dfrac{1}{2}x+\dfrac{2}{3}x-1=-3\dfrac{1}{3}\)

\(\Rightarrow x\cdot\left(\dfrac{1}{2}+\dfrac{2}{3}\right)-1=-\dfrac{10}{3}\)

\(\Rightarrow x\cdot\dfrac{7}{6}=-\dfrac{10}{3}+1\)

\(\Rightarrow x\cdot\dfrac{7}{6}=-\dfrac{7}{3}\)

\(\Rightarrow x=-\dfrac{7}{3}:\dfrac{7}{6}\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\).

Answer 2

\(\dfrac{1}{2}x+\dfrac{2}{3}x-1=-3\dfrac{1}{3}\)

=>\(x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=-\dfrac{10}{3}+1\)

=>\(x\cdot\dfrac{7}{6}=-\dfrac{7}{3}\)

=>\(\dfrac{x}{6}=-\dfrac{1}{3}\)

=>\(x=-\dfrac{6}{3}=-2\)