Question

Помогите, пожалуйста.

Answer 1

$1+log_6 frac{x+3}{x+7} =frac{1}{4}log_{sqrt6}(x-1)^2; ,\\ODZ:; ; left { {{ frac{x+3}{x+7} textgreater 0 } atop {(x-1)^2 textgreater 0}} right. ; left { {{xin (-infty ,-7)cup (-3,+infty )} atop {xne 1}} right. \\ xin (-infty ,-7)cup (-3,1)cup (1,+infty )\\log_66+log_6frac{x+3}{x+7}=frac{1}{2}log_6(x-1)^2\\log_6frac{6(x+3)}{x+7}=log_6sqrt{(x-1)^2}\\frac{6x+18}{x+7}=|x-1|\\a); x-1 geq 0; ,; x geq 1; ,; |x-1|=x-1\\frac{6x+18}{x+7}=(x-1)$

$6x+18=x^2+6x-7\\x^2-25=0; ; Rightarrow ; ; x=pm 5\\x=-5notin ODZ\\b); x-1 textless 0; ,; x textless 1; ,; |x-1|=-(x-1)\\frac{6x+18}{x+7}=-(x-1)\\6x+18=-x^2-6x+7\\x^2+12x+11=0\\x_1=-11; ,; x_2=-1$

$Otvet:; ; x=-11,; x=-1,; x=5.$