Question

f(x)=log0.5(x^2+3x)+корень(49-x^2)
f(x)=0.3^log0.3(x^2-1)

Answer 1

F(x)=log_0,5(x²+3x)+ √(49-x^2)ОДЗ х²+3х>0       x(x+3)>0       x>0                     или  х<0       x+3>0 ⇒ x>-3             x+3<0 ⇒  x<-3      x∈(-∞;-3)∪(0;+∞)
  49-x² ≥0  x²≤49   x∈[-7;7]  объединим ОДЗ х∈ [-7;-3)∪(0;7]


F(x)=0,3^log_0,3(x²-1)ОДЗx²-1>0x² >1x∈(-∞;-1)∪(1 ;+∞)

Answer 2

$1)\; \; f(x)=log_{0,5}(x^2+3x)+\sqrt{49-x^2}\\\\ \left \{ {{x^2+3x\ \textgreater \ 0} \atop {49-x^2 \geq 0}} \right. \; \left \{ {{x(x+3)\ \textgreater \ 0} \atop {(7-x)(7+x) \geq 0}} \right.\; \left \{ {{x\in (-\infty ,-3)\cup (0,+\infty )} \atop {x\in [-7,7\, ]}} \right. \\\\x\in [-7,-3)\cup (0,7\, ]\\\\2)\; \; f(x)=0,3^{log_{0,3}(x^2-1)}\\\\x^2-1\ \textgreater \ 0\\\\(x-1)(x+1)\ \textgreater \ 0\\\\x\in (-\infty ,-1)\cup (1,+\infty )$