Lg(х+√3)+lg(х-√3)=0 как решить ?

lg(x+√3)+lg(x-√3)=0 ОДЗ: [latex] \left \{ {{x+ \sqrt{3}\ \textgreater \ 0 } \atop {x- \sqrt{3} \ \textgreater \ 0}} \right. , \left \{ {{x\ \textgreater \ - \sqrt{3} } \atop {x\ \textgreater \ \sqrt{3} }} \right. =\ \textgreater \ [/latex] x>√3 lg((x+√3)*(x-√3))=0 lg(x²-(√3)²)=0 lg(x²-3)=0 x²-3=10⁰ x²-3=1 x²=4 x₁=-2, x₂=3 -2<√3, => x=-2 посторонний корень ответ: х=2

[latex]\lg(x+\sqrt3)+\lg(x-\sqrt3)=0\\O.D.3.:\;\\\begin{cases}x+\sqrt3\ \textgreater \ 0\\x-\sqrt3\ \textgreater \ 0\end{cases}\Rightarrow\begin{cases}x\ \textgreater \ -\sqrt3\\x\ \textgreater \ \sqrt3\end{cases}\Rightarrow x\ \textgreater \ \sqrt3\\\log_ab+\log_ac=\log_a(b\cdot c)\\\lg(x+\sqrt3)(x-\sqrt3)=0\\\lg(x^2-3)=0\\x^2-3=10^0\\x^2-3=1\\x^2=4\\x_1=2\\x_2=-2\;-\;He\;nogx.\;no\;O.D.3.\\OTBET:\;x=2[/latex]