a, \(\)\(\left|x+\dfrac{2}{3}\right|+2=\dfrac{7}{3}\)
⇔\(\left|x+\dfrac{2}{3}\right|=\dfrac{7}{3}-2\)
⇔\(\left|x+\dfrac{2}{3}\right|=\dfrac{1}{3}\)
⇔\(\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{3}-\dfrac{2}{3}\Rightarrow x=\dfrac{-1}{3}\\x+\dfrac{2}{3}=\dfrac{-1}{3}\Rightarrow x=\dfrac{-1}{3}-\dfrac{2}{3}\Rightarrow x=\left(-1\right)\end{matrix}\right.\)
Vậy x = \(\dfrac{-1}{3}\) hoặc x = (-1) thì \(\left|x+\dfrac{2}{3}\right|+2=\dfrac{7}{3}\)
b,
Xét \(25^{50}\) và \(5^{300}\) ta có:
\(25^{50}=\left(5^2\right)^{50}=5^{100}\)
\(5^{300}=\left(5^3\right)^{100}=125^{100}\)
Vì \(5^{100}< 125^{100}\) nên \(25^{50}< 5^{300}\)
