Question

Решите пожалуйста, фото прикрепил

Answer 1

$\displaystyle\bf\\1)\\\\3^{x} =\frac{1}{9} \\\\\\3^{x} =\frac{1}{3^{2} } \\\\\\3^{x} =3^{-2} \\\\\\\boxed{x=-2}\\\\2)\\\\3^{2x} -3^{x} -5=0\\\\3^{x} =m \ \ ; \ \ m > 0\\\\m^{2} -m-5=0\\\\D=(-1)^{2} -4\cdot(-5)=1+20=21\\\\\\m_{1} =\frac{1+\sqrt{21} }{2} \\\\\\m_{2} =\frac{1-\sqrt{21} }{2} \ < 0 \ - \ ne \ podxodit\\\\\\3^{x} =\frac{1+\sqrt{21} }{2} \\\\\\x=\log_{3} \frac{1+\sqrt{21} }{2}$

$\displaystyle\bf\\3)\\\\x^{2} -6x+5=0\\\\Teorema \ Vieta \ :\\\\x_{1} + x_{2} =6\\\\x_{1} \cdot x_{2} =5\\\\x_{1} = 1 \ \ ; \ \ x_{2} =5$