chỉ em với ạ

$\color{lightblue}{\text{o Lâm Vũ}}$

`e`) ($\dfrac{1}{2}$`x` + $\dfrac{1}{3}$) - `2,5` = $\dfrac{-7}{2}$

$\dfrac{1}{2}$`x` + $\dfrac{1}{3}$ - $\dfrac{5}{2}$ = $\dfrac{-7}{2}$

$\dfrac{1}{2}$`x` + $\dfrac{1}{3}$ = $\dfrac{-7}{2}$ + $\dfrac{5}{2}$

$\dfrac{1}{2}$`x` + $\dfrac{1}{3}$ = $\dfrac{-2}{2}$ = `-1`

$\dfrac{1}{2}$`x` = `-1` - $\dfrac{1}{3}$

$\dfrac{1}{2}$`x` = $\dfrac{-3}{3}$ - $\dfrac{1}{3}$

$\dfrac{1}{2}$`x` = $\dfrac{-4}{3}$

`x` = $\dfrac{-4}{3}$ : $\dfrac{1}{2}$

`x` = $\dfrac{-8}{3}$

Vậy `x` = $\dfrac{-8}{3}$