Question

va rog mult, repedeee​

Answer 1

Răspuns:

a) ; b)

Explicație pas cu pas:

a)

$\begin{cases} 2x - xy + 2y = 5\\ 3xy - x - y = 5\end{cases}\Big|^{(\cdot 3)} \iff \begin{cases}6x - 3xy + 6y = 15 \\3xy - x - y = 5 \end{cases}$

$5(x + y) = 20 \iff x + y = 4 \implies y = 4 - x$

$3(4 - x)x - (4 - x) - x = 5 \\ 12x - 3 {x}^{2} - 4 + x - x - 5 = 0 \\ - 3 {x}^{2} + 12x - 9 = 0 \\ {x}^{2} - 4x + 3 = 0 \\ (x - 1)(x - 3) = 0$

$\bf x_{1} = 1 \implies y_{1} = 3 \\ \bf x_{2} = 3 \implies y_{2} = 1$

.

b) x ≠ 0, y ≠ 0

$\begin{cases} \frac{1}{x} + \frac{1}{y} = 4 \\ {x}^{2} + {y}^{2} = \frac{1}{2} \end{cases} \iff \begin{cases} \frac{x + y}{xy} = 4 \\ 2{x}^{2} + 2{y}^{2} - 1 = 0 \end{cases}$

$\begin{cases} x + y = 4xy \\ 2{(x + y)}^{2} - 4xy - 1 = 0 \end{cases} \iff \begin{cases} x + y = 4xy \\ 2{(x + y)}^{2} - (x + y) - 1 = 0 \end{cases}$

$\begin{cases} x + y = 4xy \\ (2x + 2y + 1)(x + y - 1) = 0 \end{cases}$

I)

$x + y - 1 = 0 \implies y = 1 - x$

$2 {x}^{2} + 2 {(1 - x)}^{2} - 1 = 0 \\ 4 {x}^{2} - 4x + 1 = 0 \iff {(2x - 1)}^{2} = 0 \\ \bf x_{1} = \frac{1}{2} \implies y_{1} = \frac{1}{2}$

II)

$2x + 2y + 1 = 0 \implies y = - \frac{2x + 1}{2} \iff y = - x - \frac{1}{2}$

$2 {x}^{2} + 2 {\Big(- \frac{2x + 1}{2}\Big)}^{2} - 1 = 0$

$8 {x}^{2} + 8 {x}^{2} + 8x + 2 - 4 = 0 \\ 8 {x}^{2} + 4x - 1 = 0$

$\Delta = 16 + 32 = 48$

$x_{1;2} = \frac{ - 4 \pm 4\sqrt{3} }{16} = \frac{ - 1 \pm \sqrt{3} }{4}$

$\bf x_{2} = - \frac{1 + \sqrt{3} }{4} \implies y_{2} = - \frac{1 - \sqrt{3} }{4} \\ \bf x_{3} = - \frac{1 - \sqrt{3} }{4} \implies y_{3} = - \frac{1 + \sqrt{3} }{4}$