Question

y=1+x
x+y в квадрате = - 1

Answer 1

$y = 1 + x$
$x + {y}^{2} = - 1$

$x + {(1 + x)}^{2} = - 1$

$x + 1 + 2x + {x}^{2} + 1 = 0$

${x}^{2} + 3x + 2 = 0$

$D = 9 - 4 \times 2 = 1$

$\sqrt{D} = 1$

$x1 = \frac{ - 3 - 1}{2} = - 2$

$x2 = \frac{ - 3 + 1}{2} = - 1$

Ответ: х1 = -2; х2 = -1

Answer 2

$y = 1 + x \\ {(x + y)}^{2} = - 1 \\ {x}^{2} + {y}^{2} + 2xy = - 1 \\ {x}^{2} + {(1 + x)}^{2} + 2x (1 + x) = - 1 \\ {x}^{2} + 1 + {x}^{2} + 2x + 2 {x}^{2} = - 1 \\ 4 {x}^{2} + 2x + 2 = 0$
D=-4-4*4*2=-4-32=-36
$x1 = \frac{ - 2 + 36}{4} = \frac{34}{4} = \frac{17}{2} = 8.5$
$x2 = \frac{ - 2 - 36}{4} = \frac{ - 38}{4} = \frac{19}{2} = 9.5$