Question

[x²-2y² +5xy +2=0
2³x-y = 1 ​

Answer 1

Ответ:

${x}^{2} - 2 {y}^{2} + 5xy + 2 = 0 \\ {2}^{3} x - y = 1$

${x}^{2} - 2 {y}^{2} + 5xy + 2 = 0 \\ 8x - y = 1$

$8x - y = 1 \\ - y = 1 - 8x \\ y = - 1 + 8x$

${x}^{2} - 2 {y}^{2} + 5xy + 2 = 0 \\ y = - 1 + 8x$

${x}^{2} - 2( - 1 + 8x) ^{2} + 5x \times ( - 1 + 8x) + 2 = 0$

${x}^{2} - 2(8x - 1) ^{2} - 5x + 4 {x}^{2} + 2 = 0$

${x}^{2} - 2(64 {x}^{2} - 16x + 1) - 5x + 40 {x}^{2} + 2 = 0$

${x}^{2} - 128 {x}^{2} + 32x - 2 - 5x + 40 {x}^{2} + 2 = 0$

$- 87 {x}^{2} + 27x = 0$

$- 3x \times (29x - 9) = 0$

$\div - 3$

$x \times (29x - 9) = 0$

$x = 0 \\ 29x - 9 = 0$

$x = 0 \\ x = \frac{9}{29}$

$y = - 1 + 8 \times 0 \\ y = - 1 + 8 \times \frac{9}{29}$

$y = - 1 + 8 \times 0 \\ y = - 1 + 0 \\ y = - 1$

$y = - 1 + 8 \times \frac{9}{29} \\ y = - 1 + \frac{72}{29 } \\ y = \frac{43}{29}$

$y = - 1 \\ y = \frac{43}{29}$

$( x_{1}. x_{2}) = (0. - 1) \\ ( x_{2}. y_{2}) = ( \frac{9}{29} . \frac{43}{29} )$

${0}^{2} - 2 \times ( - 1) ^{2} + 5 \times 0 \times ( - 1) + 2 = 0 \\ {2}^{3} \times 0 - ( - 1) = 1 \\ \\ ( \frac{9}{29} ) ^{2} - 2 \times ( \frac{43}{29} ) ^{2} + 5 \times \frac{9}{29} \times \frac{43}{29} + 2 = 0 \\ {2}^{3} \times \frac{9}{29} - \frac{43}{29} = 1$

$0 = 0 \\ 1 = 1 \\ \\ 0 = 0 \\ 1 = 1$

$( x_{1}. x_{2}) = (0. - 1) \\ ( x_{2}. y_{2}) = ( \frac{9}{29} . \frac{43}{29} )$