Question

Пожалуйста решите уравнение срочно (√7)^2х+2-50*(√7)^х+7=0

Answer 1

$(\sqrt{7})^{2x+2}-50*(\sqrt{7})^{x}+7=0\\\\(\sqrt{7})^{2x}*(\sqrt{7})^{2}-50*(\sqrt{7})^{x}+7=0\\\\(\sqrt{7} )^{x}=m,m>0\\\\7m^{2} -50m+7=0\\\\D=(-50)^{2}-4*7*7=2500-196=2304=48^{2}\\\\m_{1}=\frac{50+48}{14}=7\\\\m_{2}=\frac{50-48}{14}=\frac{1}{7}=7^{-1}\\\\(\sqrt{7})^{x}=7\\\\7^{\frac{1}{2}x }=7\\\\\frac{1}{2}x=1\\\\x_{1}=2\\\\(\sqrt{7})^{x}=7^{-1}\\\\7^{\frac{1}{2}x }=7^{-1}\\\\\frac{1}{2}x=-1\\\\x_{2}=-2$