Решить , решение полное надо с одз и т.д
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ОДЗ:
$7 - frac{x}{2} textgreater 0; Rightarrow - frac{x}{2} textgreater -7 Rightarrow frac{x}{2} textless 7 Rightarrow x textless 14 22-x textgreater 0 Rightarrow -x textgreater -22 Rightarrow x textless 22 boxed{x textless 14} frac{log_{5^2}(7 - frac{x}{2})}{log_{5^3} (22-x)} leq frac{3}{4}; frac{frac{1}{2} cdot log_5(7 - frac{x}{2})}{frac{1}{3} cdot log_5 (22-x)} leq frac{3}{4}; frac{3 cdot log_5(7 - frac{x}{2})}{2 cdot log_5 (22-x)} leq frac{3}{4};$

$frac{3 cdot log_5(7 - frac{x}{2})}{2 cdot log_5 (22-x)} leq frac{3}{4}; frac{log_5(7 - frac{x}{2})}{ log_5 (22-x)} leq frac{1}{2}; frac{2cdot log_5(7 - frac{x}{2})}{ log_5 (22-x)} leq 1; frac{2cdot log_5(frac{1}{2} cdot (14 - x))}{ log_5 (22-x)}-1 leq 0; frac{2cdot (log_5frac{1}{2} + log_5(14 - x))-log_5 (22-x)}{ log_5 (22-x)} leq 0; frac{log_5frac{1}{4} + log_5(14 - x)^2-log_5 (22-x)}{ log_5 (22-x)} leq 0;$

$log_5 (22-x) neq 0 Rightarrow log_5 (22-x) neq log_5 1 Rightarrow 22-x neq 1 x neq 21$

$log_5frac{1}{4} + log_5(14 - x)^2-log_5 (22-x) = 0 log_5(frac{(14-x)^2}{4 cdot (22-x)})=0 log_5(frac{196 -28x+x^2}{88 -4x})=log_5 1 frac{196 -28x+x^2}{88 -4x}=1; frac{196 -28x+x^2}{88 -4x}-1=0; frac{196 -28x+x^2-88+4x}{88 -4x}=0 x^2-24x +108=0; x_{1,2}=frac{24 pm sqrt{576-342}}{2}=frac{24 pm 12}{2}; x_1=18, x_2=6;$

$88-4x neq 0 Rightarrow 4x neq 88 Rightarrow x neq 22$

+ — + — +
--------------*----------------*-------------o------------o-------->x
6 18 21 22

$x in [6; 18] cup (21;22)$

C учетом ОДЗ:
$x textless 14; 6 leq x leq 18; 21<x<22 boxed{6 leq x textless 14 RIghtarrow x in [6;14)}$