Question

найдите сумму целых решений

Answer 1

$log_{25}(\sqrt7-2)\cdot log_5(3+x)^2\ \textgreater \ \frac{1}{log_{\sqrt7-2}5} \\\\ODZ:\; \; (3+x)^2\ \textgreater \ 0\; \; \Rightarrow \; \; 3+x\ne 0\; ,\; \underline {x\in (-\infty ,-3)\cup (-3,+\infty )}\\\\ \frac{1}{2}\cdot log_5(\sqrt7-2)\cdot log_5(3+x)^2\ \textgreater \ log_5(\sqrt7-2)\, \Big |:log_5(\sqrt7-2)\ \textless \ 0\\\\\star \; \sqrt7-2\approx 0,65\; \; \Rightarrow \; \; log_5(\sqrt7-2)\approx log_5\, 0,65\ \textless \ 0\; \; [\, log_55=1\, ]\; \star \Rightarrow \\\\ \frac{1}{2}\cdot log_5(3+x)^2\ \textless \ 1\\\\log_5(3+x)^2\ \textless \ 2\; \; \Rightarrow \; \; log_5(3+x)^2\ \textless \ log_55^2\\\\(3+x)^2\ \textless \ 5^2\; \; \Rightarrow \; \; (3+x-5)(3+x+5)<0$

$(x-2)(x+8)\ \textless \ 0\; \; \; \to \; \; x_1=2\; ,\; \; x_2=-8\\\\+++(-8)---(2)+++\\\\x\in (-8,2)\\\\ \left \{ {{x\in (-8,2)} \atop {x\in (-\infty ,-3)\cup (-3,+\infty )}} \right. \; \; \; \Rightarrow \; \; \; \underline {x\in (-8,-3)\cup (-3,2)}\\\\Celue\; resheniya:\; \; x=-7,-6,-5,-4,-2,-1,0,1.\\\\Symma\; celux\; reshenij\; =\; -7-6-5-4-2-1+0+1=-24.$