Question

Помогите решить по математике ​

Answer 1

$\frac12\lg\left(x^2-4x-1\right)=\lg(8x)-\lg(4x)\\O.D.3.:\\\begin{cases}x^2-4x-1>0\\8x>0\\4x>0\end{cases}\Rightarrow\begin{cases}(x-2+\sqrt5)(x-2-\sqrt5)>0\\x>0\end{cases}\Rightarrow\\\Rightarrow x\in(2+\sqrt5;\;+\infty)\\\\\lg\left(x^2-4x-1\right)^{\frac12}=\lg(8x:4x)\\\\\sqrt{x^2-4x-1}=2\\x^2-4x-1=4\\x^2-4x-5=0\\D=16-4\cdot1\cdot(-5)=16+20=36\\x_{1,2}=\frac{4\pm6}2\\x_1=-1\;-\;He\;nogx.\;no\;O.D.3.\\x_2=5$