Tìm x, biết rằng: 1. 4x^2 +1 = 8 - căn bậc hai của 6 2. x^2 +1

* Hướng dẫn giải

1.

$\begin{array}{l}4x^2+1=8-2\sqrt{6}\\ \text{⇔}\left(2x^2\right)=7-2\sqrt{6}\\ \text{⇔}\left(2x^2\right)={\left(\sqrt{6}-1\right)}^2\\ \left[\begin{array}{l}x=\frac{\sqrt{6}-1}{2}\\ x=\frac{1-\sqrt{6}}{2}\end{array}\right.\end{array}$

Vậy $S=\left\{\frac{1-\sqrt{6}}{2};\frac{\sqrt{6}-1}{2}\right\}$

2. $x^2+1=6-2\sqrt{6}$

$\begin{array}{l}\iff x^2=5-2\sqrt{6}\\ \iff x^2={\left(\sqrt{3}-\sqrt{2}\right)}^2\\ \iff \left[\begin{array}{l}x=\sqrt{3}-\sqrt{2}\\ x=\sqrt{2}-\sqrt{3}\end{array}\right.\end{array}$

Vậy $S=\left\{\sqrt{3}-\sqrt{2};\sqrt{2}-\sqrt{3}\right\}$