Question

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Дуже потрібно до завтра

Answer 1

Ответ:

$1) \ \ (x; \ y) = (2; \ 4)$

$2)$  нет ответа

Объяснение:

№1

$\begin{cases}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\\\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{4}\end{cases}\\\\\\\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{3}{4}+\dfrac{1}{4}\\\\\\2\times \dfrac{1}{x}=\dfrac{4}{4}\\\\\\ \dfrac{2}{x}=1\\\\\\\boxed{x=2}\\\\\\\dfrac{1}{2} - \dfrac{1}{y}=\dfrac{1}{4}\\\\\\\dfrac{1}{y}=\dfrac{1}{2} - \dfrac{1}{4}\\\\\\\dfrac{1}{y}=\dfrac{2}{4} - \dfrac{1}{4}\\\\\\\dfrac{1}{y}=\dfrac{1}{4}\\\\\\\boxed{y=4}$

    $\downarrow$       $\downarrow$

$\boxed{(x; \ y) = (2; \ 4)}$

№2

$\begin{cases}1+\dfrac{x}{x+1}=\dfrac{y}{1-x^2}\\\\\dfrac{x-5}{3-y}=\dfrac{1}{2}\end{cases}\\\\\\\begin{cases}1+\dfrac{x}{x+1}=\dfrac{y}{1-x^2}\\\\2(x-5)=3-y\end{cases}\\\\\\\begin{cases}1+\dfrac{x}{x+1}=\dfrac{y}{1-x^2}\\\\2x-10=3-y\end{cases}\\\\\\\begin{cases}1+\dfrac{x}{x+1}=\dfrac{y}{1-x^2}\\\\y=13-2x\end{cases}\\\\\\1+\dfrac{x}{x+1} =\dfrac{13-2x}{1-x^2}\\\\\\1+\dfrac{x}{1+x} =\dfrac{13-2x}{(1-x)(1+x)}\\\\\\(1-x)(1+x)+x(1-x)=13-2x\\\\1-x^2+x-x^2=13-2x\\\\-2x^2+3x-12=0$

$=================\\\\D=b^2-4ac\\\\a=-2; \quad b=3; \quad c=-12;\\\\D=3^2-4 \times(-2) \times (-12)\\\\D=9-96\\\\D=-87\\\\=================$

$D < 0$  —  нет ответа