Question

Терміново потрібно виконати

Answer 1

Ответ:

Выразим переменную у через переменную х .

$\bf1 )\ \ x^2+y^2=1\ \ \ \Rightarrow \ \ y^2=1-x^2\ \ ,\ \ \underline{y=\pm \sqrt{1-x^2}\ ,\ -1\leq x\leq 1\ ,\ -1\leq y\leq 1}\\\\\\2)\ \ xy=a\ \ \ \Rightarrow \ \ \ \underline{y=\dfrac{a}{x}\ ,\ x\ne 0}\\\\\\3)\ \ 2^{xy}=5\ \ \ \Rightarrow \ \ \ xy=log_25\ \ ,\ \ \underline{y=\dfrac{log_25}{x}\ ,\ x\ne 0}\\\\\\4)\ \ lnx+ln(y+1)=4\ \ \ \Rightarrow \ \ ln\Big(x(y+1)\Big)=ln\, e^4\ \ ,\ \ x(y+1)=e^4\ ,\\\\y+1=\dfrac{e^4}{x}\ \ ,\ \ \underline{y=\dfrac{e^4}{x}-1\ ,\ x > 0\ ,\ y > -1}$

$\bf 5)\ \ 2^{x+y}(x^2-2)=x^3+7\ \ \Rightarrow \ \ 2^{x}\cdot 2^{y}(x^2-2)=x^3+7\ \ ,\\\\2^{y}=\dfrac{x^3+7}{2^{x}(x^2-2)}\ \ ,\ \ \underline{y=log_2\, \dfrac{x^3+7}{2^{x}(x^2-2)}\ ,\ x\in (-\sqrt[3]7;-\sqrt2)\cup (\sqrt2;+\infty )}\\\\\\6)\ \ (1+x)cosy-x^2=0\ \ \Rightarrow \ \ cosy=\dfrac{x^2}{1+x}\ ,\\\\\underline{y=\pm arccos\dfrac{x^2}{1+x}+2\pi n\ ,\ \ n\in Z\ ,\ \ 0\leq \dfrac{x^2}{1+x}\leq \pi }$