Question

\(a.\dfrac{-4}{3}+a^2-4=\dfrac{7}{3}\)

Tìm a 

Answer 1

\(\Leftrightarrow a^2-\dfrac{4}{3}a-\dfrac{19}{3}=0\)

\(\Leftrightarrow3a^2-4a-19=0\)

\(\text{Δ}=\left(-4\right)^2-4\cdot3\cdot\left(-19\right)=244\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}a_1=\dfrac{4-2\sqrt{61}}{6}=\dfrac{2-\sqrt{61}}{3}\\a_2=\dfrac{2+\sqrt{61}}{3}\end{matrix}\right.\)

Answer 2

\(\dfrac{-4}{3}+a^2-4=\dfrac{7}{3}\) ⇔ \(a^2=\dfrac{4}{3}+4+\dfrac{7}{3}\) ⇔\(a^2=\dfrac{11}{3}+4\) ⇔\(a^2=\dfrac{23}{3}\) ⇔ \(\left[{}\begin{matrix}a=\sqrt{\dfrac{23}{3}}\\a=-\sqrt{\dfrac{23}{3}}\end{matrix}\right.\)