Question

все! пожалуйста! спасибо!​(на листочке бумажном с ручкой)

Answer 1

Ответ:

$8)\ \ tg\Big(\dfrac{3\pi}{2}-x\Big)=1\\\\\dfrac{3\pi}{2}-x=\dfrac{\pi}{4}+\pi n\ ,\ n\in Z\\\\x=\dfrac{3\pi}{2}-\dfrac{\pi}{4}-\pi n\ ,\ n\in Z\\\\x=\dfrac{5\pi}{4}-\pi n\ ,\ n\in Z\\\\x\in [\ 0\ ;\frac{\pi}{2}\ ]\ \ \Rightarrow \ \ x=\dfrac{5\pi}{4}-\pi =\dfrac{\pi }{4}=\boxed{\ 45^\circ \ }$

$9)\ \ x=2\pi \\\\1+ctg\Big(\dfrac{3\pi}{2}+x\Big)\cdot sinx\cdot cosx=1+tgx\cdot sinx\cdot cosx=1+\dfrac{sinx}{cosx}\cdot sinx\cdot cosx=\\\\=1+sin^2x=1+sin^2(2\pi )=1+0^2=1\\\\\\10)\ \ 5^{log_5(x-2)}=x^2-4x+4\ \ ,\ \ \ \ \ ODZ:\ x-2>0\ \ ,\ \ x>2\ ,\\\\(x-2)=(x-2)^2\ \ ,\ \ (x-2)^2-(x-2)=0\ \ ,\ \ \ (x-2)(x-2-1)=0\ ,\\\\(x-2)(x-3)=0\ \ \Rightarrow \ \ \ x_1=2\notin ODZ\ ,\ \ \ x_2=3\in ODZ\\\\Otvet:\ x=3\ .$

$11)\ \ \Big(4,3\sqrt[5]{5\sqrt{5\sqrt5}}+0,7\sqrt[20]{5^7}\Big)^{\frac{20}{27}}=\Big(4,3\cdot 5^{\frac{7}{20}}+0,7\cdot 5^{\frac{7}{20}}\Big)^{\frac{20}{27}}=\\\\\\=\Big(5\cdot 5^{\frac{7}{20}}\Big)^{\frac{20}{27}}=\Big(5^{\frac{27}{20}}\Big)^{\frac{20}{27}}=5\\\\\\\star \ \ \sqrt[5]{5\sqrt{5\sqrt5}}=\sqrt[5]{\sqrt{5^3\sqrt5}}=\sqrt[10]{\sqrt{5^7}}=\sqrt[20]{5^7}=5^{\frac{7}{20}}\ \ \star$

$12)\ \ x=\sqrt{4+2x-x^2}+2\ \ ,\\\\\\ODZ:\ \left\{\begin{array}{l}x>0\\4+2x-x^2\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x>0\\x^2-2x-4\leq 0\end{array}\right\ \ \left\{\begin{array}{l}x\in (\ 0\ ;+\infty )\\x\in [\ 1-\sqrt5\ ;\ 1+\sqrt5\ ]\end{array}\right\\\\\\x\in (\ 0\ ;\ 1+\sqrt5\ ]\\\\\\\sqrt{4+2x-x^2}=x-2\ \ ,\ \ \ x-2\geq 0\ ,\ x\geq 2\ ,\\\\4+2x-x^2=x^2-4x+4\ \ ,\ \ \ 2x^2-6x=0\ \ ,\ \ \ 2x\, (x-3)=0\ \ ,\\\\x_1=0\notin ODZ\ \ ,\ \ x_2=3\in ODZ$

$Proverka:\ \ x=3:\ \ \sqrt{4+2\cdot 3-3^2}+2=\sqrt{1}+2=1+2=3\ ,\ \ 3=3\ \ .\\\\Otvet:\ x=3\ .$