Question

Lg(x-6)-0,5lg2=lg3+lgкорень квадратный из x-10

Answer 1

$lg(x-6)-0,5lg2=lg3+lgsqrt{x-10}; ,\\ ODZ:; ; left { {{x-6 textgreater 0} atop {x-10 textgreater 0}} right. ; left { {{x textgreater 6} atop {x textgreater 10}} right. ; ,; x textgreater 10\\lgfrac{x-6}{sqrt 2}=lg(3sqrt{x-10})\\frac{x-6}{sqrt 2}=3sqrt{x-10}\\sqrt{x-10}cdot sqrt2=frac{x-6}{3}\\2(x-10)=frac{x^2-12x+36}{9} \\18x-180=x^2-24x+72\\x^2-42x+252=0$

$D/4=21^2-252=189 ,\\ x_1=21-sqrt{189}=21-3sqrt{21} textless 10; ,; x_2=21+3sqrt{21} textgreater 10\\Otvet:; ; x=21+3sqrt{21}=sqrt{21}cdot (sqrt{21}+3); .$