Giải bất phương trình căn bậc 2(x^2 + x - 12) < 6 - x.

* Hướng dẫn giải

$\sqrt{x^2+x-12}<6-x\iff \left\{\begin{array}{l}6-x>0\\ x^2+x-12\ge 0\\ x^2+x-12<{\left(6-x\right)}^2\end{array}\right.$

$\iff \left\{\begin{array}{l}x<6\\ x\in \left(-\infty ;-4\right]\cup \left[3;+\infty \right)\\ x^2+x-12<36-12x+x^2\end{array}\right.$

$\iff \left\{\begin{array}{l}x<6\\ x\in \left(-\infty ;-4\right]\cup \left[3;+\infty \right)\\ 13x<48\end{array}\right.$

$\iff \left\{\begin{array}{l}x<6\\ x\in \left(-\infty ;-4\right]\cup \left[3;+\infty \right)\\ x<\frac{48}{13}\end{array}\right.$

$\iff x\in \left(-\infty ;-4\right]\cup \left[3;\frac{48}{13}\right)$