Question

Помогите решить до 10.11.2021

Answer 1

$4x {}^{2} - 3x = y \\ 8x - 6 = y$

т.к. левая часть уравнений = y, то можно их приравнять

$4x {}^{2} - 3x = 8x - 6 \\ 4x {}^{2} - 3x - 8x+ 6 = 0$

$4x { }^{2} - 11x + 6 = 0$

$d = b {}^{2} - 4ac$

$d = 11 {}^{2} - 4 \times 4 \times 6 = 121 - 96 = 25$

$\sqrt{25} = 5$

$x1x2 = \frac{ - b + - \: \sqrt[]{d} }{2a}$

$x1x2 = \frac{ - ( - 11) + - \sqrt{25} }{2 \times 4}$

$x1 = \frac{11 - 5}{8} = \frac{6}{8} = \frac{3}{4} = 0.75$

$x2 = \frac{11 + 5}{8} = \frac{16}{8} = 2$

$y1 = 8 \times x1 - 6$

$y2 = 8 \times x2 - 6$

$y1 = 8 \times 0.75 - 6 = 6 - 6 = 0$

$y2 = 8 \times 2 - 6 = 16 - 6 = 10$

Ответ: ( x1; y1) = ( 0.75; 0)

( x2; y2) = ( 2; 10)

$(2x + 4) {}^{2} = 3y \\ (4x + 2) ^{2} = 3y$

также приравниваем

$(2x + 4) {}^{2} = (4x + 2) {}^{2} \\ 4x {}^{2} + 16x + 16 = 16x {}^{2} + 16x+ 4$

$16x {}^{2} + 16x + 4 - 4x {}^{2} - 16x - 16 = 0$

$12x {}^{2} - 12 = 0$

обе части делим на 12

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

$x1 = - 1 \\ x2 = 1$

$3y 1 = (4 \times x1 + 2) {}^{2}$

$3y2 = (4 \times x2 + 2) {}^{2}$

$3y1 =( 4 \times ( - 1) + 2) {}^{2} \\ 3y1 = ( - 4 +2 ) {}^{2} \\ 3y1 = ( - 2) {}^{2} \\ 3y1 = 4 \\ y = \frac{3}{4} = 0.75$

$3y2 = ( 4\times 1 + 2) {}^{2} \\ 3y2 =( 4 + 2) {}^{2} \\ 3y2 = 6 { }^{2} \\ 3y2 = 36 \\ y = 12$

Ответ: ( x1; y1) = ( -1; 0.75)

( x2; y2) = ( 1; 12)

Очень подробно рассписал, надеюсь помог