Ответ:
ОДЗ :
x + 2 > 0
x > - 2
log_{4}(x+2)+log_{4}3=log_{4}15\\\\log_{4}(x+2)=log_{4}15-log_{4}3\\\\log_{4} (x+2)=log_{4}\frac{15}{3}\\\\log_{4}(x+2)=log_{4}5\\\\x+2=5\\\\x=3\\\\Otvet:3
Объяснение:
<span>1) (7-х)(7+х)+(х+3)</span>² = 7² - х² + (х² + 2·х·3 + 9) = 49 - х² + х² + 6х + 9 =
= 58 + 6х = 2·(29 +3х)
<span>
2) (12р</span>⁶<span>q</span>²<span>)/(8p</span>⁴<span>q</span>⁵) = (3p²)/(2q³)<span> </span>