Question

АЛГЕБРА 10-11 КЛАСС СРОЧНО! 100 БАЛЛОВ!!!

Answer 1

$4)\ \ 3^{2x}-9\cdot 3^{x}+8=0\\\\t=3^{x}>0\ ,\ \ \ t^2-9t+8=0\ \ ,\ \ t_1=1\ ,\ t_2=8\ \ (teorema\ Vieta)\\\\3^{x}=1\ \ ,\ \ 3^{x}=3^0\ \ \to \ \ \ x=0\\\\3^{x}=8\ \ ,\ \ 3^{x}=3^{log_38}\ \ \to \ \ \ x=log_38\\\\Otvet:\ \ x_1=0\ ,\ x_2=log_38\ .$

$5)\ \ y=3x^3-9x^2-1\\\\y'=9x^2-18x=9x\, (x-2)=0\ \ ,\ \ x_1=0\ ,\ x_2=2\\\\znaki\ y':\ \ \ +++(0)---(2)+++\\\\{}\qquad \qquad \ \ \ \quad \nearrow \ \ \ (0)\ \ \searrow \ \ \ (2)\ \ \nearrow \\\\x_{max}=0\ \ ,\ \ y_{max}=-1$

$6)\ \ 2sin^2x+3cosx-3=0\\\\2(1-cos^2x)+3cosx-3=0\ \ ,\ \ \ 2-2cos^2x+3cosx-3=0\ \ ,\\\\2cos^2x-3cosx+1=0\ \ ,\ \ D=9-8=1\ ,\\\\ cosx=\dfrac{1}{2}\ \ \ \ ili\ \ \ \ cosx=1\\\\x=\pm \dfrac{\pi }{3}+2\pi n\ ,\ n\in Z\ \ \ ili\ \ \ x=2\pi k\ ,\ k\in Z\\\\x\in [\ 0\, ;\, 2\pi \ ]:\ \ x_1=\dfrac{\pi}{3}\ ,\ x_2=\dfrac{5\pi}{3}\ \ ,\ \ x_3=2\pi \\\\Otvet:\ \ a)\ x=\pm \dfrac{\pi }{3}+2\pi n\ ,\ \ x=2\pi k\ ,\ n.k\in Z\ ;\\\\b)\ \ [\, 0\, ;\, 2\pi \ ]:\ x_1=\dfrac{\pi}{3}\ ,\ x_2=\dfrac{5\pi}{3}\ \ ,\ \ x_3=2\pi$