Đặt \(\sqrt{a^3+a^2}=t\ge0\), PTTT:
\(\sqrt{t^2+4}+\sqrt{t^2-3}=7\\ \Leftrightarrow\left(\sqrt{t^2+4}-4\right)+\left(\sqrt{t^2-3}-3\right)=0\\ \Leftrightarrow\dfrac{t^2-12}{\sqrt{t^2+4}+4}+\dfrac{t^2-12}{\sqrt{t^2-3}+3}=0\\ \Leftrightarrow\left(t^2-12\right)\left(\dfrac{1}{\sqrt{t^2+4}+4}+\dfrac{1}{\sqrt{t^2-3}+3}\right)=0\)
Dễ thấy ngoặc lớn luôn lớn hơn 0
Do đó \(t^2-12=0\Leftrightarrow t^2=12\)
\(\Leftrightarrow a^3+a^2=12\\ \Leftrightarrow a^3-2a^2+3a^2-6a+6a-12=0\\ \Leftrightarrow\left(a-2\right)\left(a^2+3a+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2\\a^2+3a+6=0\left(vô.n_0\right)\end{matrix}\right.\)
Vậy PT có nghiệm a=2
