Question

(√х+5 -√х+4 )^х^2 =(√х+5 +√х+4)^5х-6​

Answer 1

$x_1 = -4,\,\,x_2 = -3,\,\,x_3=-2$

Пошаговое объяснение:

$(\sqrt{x + 5} - \sqrt{x + 4})^{x^{2}} = (\sqrt{x + 5} + \sqrt{x + 4})^{5x+6}\\\\(\sqrt{x + 5} - \sqrt{x + 4})^{x^{2}}(\sqrt{x + 5} + \sqrt{x + 4})^{x^{2}} =\\= (\sqrt{x + 5} + \sqrt{x + 4})^{5x+6} (\sqrt{x + 5} + \sqrt{x + 4})^{x^{2}}\\\\((\sqrt{x + 5} - \sqrt{x + 4})(\sqrt{x + 5} + \sqrt{x + 4}))^{x^{2}} = (\sqrt{x + 5} + \sqrt{x + 4})^{x^{2} +5x+6}\\\\((x+5) - (x+4))^{x^{2}} = (\sqrt{x + 5} + \sqrt{x + 4})^{x^{2} +5x+6}\\\\1 = (\sqrt{x + 5} + \sqrt{x + 4})^{x^{2} +5x+6}$

$1) \sqrt{x + 5} + \sqrt{x + 4} = 1\\\\\sqrt{x + 5} = 1 - \sqrt{x + 4}\\\\x+5 = 1 - 2\sqrt{x + 4}+x+4\\\\2\sqrt{x + 4} = 0\\\\x+4 =0\\\\x = -4,\,\,OD3:x \geq -4\\\\2) x^2+5x+6 = 0\\\\D=25-4\cdot6=1\\\\x_{1,2}=\dfrac{-5\pm1}{2},\,\,x_1=-2,\,\,x_2=-3,\,\,OD3:x\geq-4$