Решите 2 любых 20 баллов !СРОЧНО !!!!!!

$3^{x^2-17x+63,5}=27\sqrt3}\\\\27\sqrt3=3^3\cdot 3^{\frac{1}{2}}=3^{3,5}\\\\x^2-17x+63,5=3,5\\\\x^2-17x+60=0\\\\D=49,\\\\x_1=\frac{17-7}{2}=5,\; \; x_2=12$

1,\; \; \; 1=2^0\\\\x^2-7x+12>0\\\\x_1=3,\; x_2=4\; (teor.\; Vieta)\\\\+++(3)---(4)+++\\\\x\in (-\infty,3\, )U(\, 4,+\infty)" alt="2^{x^2-7x+12}>1,\; \; \; 1=2^0\\\\x^2-7x+12>0\\\\x_1=3,\; x_2=4\; (teor.\; Vieta)\\\\+++(3)---(4)+++\\\\x\in (-\infty,3\, )U(\, 4,+\infty)" align="absmiddle" class="latex-formula">