Question

Помогите решить уравнение.

Answer 1

$\sqrt{7 - x^{2}} + \sqrt{6 + x^{2}} = 5\\\sqrt{7 - x^{2}} = 5 - \sqrt{6 + x^{2}}\\(\sqrt{7 - x^{2}})^{2} = (5 - \sqrt{6 + x^{2}})^{2}\\7 - x^{2} = 25 - 10\sqrt{6 + x^{2}} + 6 + x^{2}\\2x^{2} + 24 = 10\sqrt{6 + x^{2}} \ \ \ |:2\\x^{2} + 12 = 5\sqrt{6 + x^{2}}\\(x^{2} + 12)^{2} =(5\sqrt{6 + x^{2}})^{2}\\x^{4} + 24x^{2} + 144 = 150 +25x^{2}\\x^{4} - x^{2} - 6 = 0\\\left\{\begin{matrix}x_{1}^{2} + x^{2}_{2} = 1 & & \\ x_{1}^{2} \cdot x^{2}_{2} = -6 & &\end{matrix}\right.$

$\left[\begin{array}{ccc}x_{1}^{2} = 3 \ \ \\\ \\x_{2}^{2} =-2\\\end{array}\right\\\\\left[\begin{array}{ccc}x_{1} = \pm \sqrt{3} \ \ \\\ \\x_{2} = \oslash \ \ \ \ \ \\\end{array}\right$

Проверка:

$1) \ x = -\sqrt{3}: \ \sqrt{7 - (-\sqrt{3})^{2}} + \sqrt{6 + (-\sqrt{3})^{2}} = \sqrt{7 - 3} + \sqrt{6 + 3} = \sqrt{4} + \sqrt{9} = \\= 2 + 3 = 5;\\\\2) \ x = \sqrt{3} : \ \sqrt{7 - (\sqrt{3})^{2}} + \sqrt{6 + (\sqrt{3})^{2}} = \sqrt{7 - 3} + \sqrt{6 + 3} = \sqrt{4} + \sqrt{9} = \\= 2 + 3 = 5.$

Ответ: $\ \pm \sqrt{3}$