+) Nếu $x≥\dfrac{-7}{2}$ thì: $2x+7≥0$
$⇒\sqrt[]{(2x+7)^2}=|2x+7|=2x+7$
Khi đó: $2x-\sqrt[]{(2x+7)^2}=2x-(2x+7)=-7$
+) Nếu $x<\dfrac{-7}{2}$ thì: $2x+7<0$
$⇒\sqrt[]{(2x+7)^2}=|2x+7|=-2x-7$
Khi đó: $2x-\sqrt[]{(2x+7)^2}=2x-(-2x-7)=4x+7$
Vậy $2x-\sqrt[]{(2x+7)^2}=\begin{cases} -7\text{ nếu $x≥\dfrac{-7}{2}$ }\\4x+7 \text{ nếu $x<\dfrac{-7}{2}$}\end{cases}$
