Question

Решите уравнения:
1)x2/2-x=3x/2-x
2) y+4/y+2=2y-1/y
3) 5x-2/x+2=6x-21/x-3

Answer 1

Ответ:

Объяснение:

1) $\frac{x^{2}}{2-x}=\frac{3x}{2-x}/*(2-x)$

x² = 3x

x² - 3x = 0

x(x - 3) = 0

x₁ = 0

x₂ = 3

$2)\frac{y+4}{y+2}=\frac{2y-1}{y}\\\\\frac{y(y+4)-(y+2)(2y-1)}{y(y+2)}=0/*y(y+2)$

y(y + 4) - (y + 2)(2y - 1) = 0

y² + 4y - (2y² - y + 4y - 2) = 0

y² + 4y - 2y² + y - 4y + 2 = 0

y² + y + 2 = 0

D = b² - 4ac = 1 - 4*2 = -7

D < 0 корней нет

$3)\frac{5x-2}{x+2}=\frac{6x-21}{x-3}\\\\\frac{(x-3)(5x-2)-(x+2)(6x-21)}{(x+2)(x-3)}=0/*(x+2)(x-3)$

(x - 3)(5x - 2) - (x + 2)(6x - 21) = 0

5x² - 2x - 15x + 6 - (6x² - 21x + 12x - 42) = 0

5x² - 2x - 15x + 6 - 6x² + 21x - 12x + 42 = 0

-x² - 8x + 48 = 0

D = b² - 4ac = 64 - 4*(-1)*48 = 256

√D = 16

$x_{1,2}=\frac{-b+-\sqrt{D}}{2a}$

x₁ = (8 + 16)/(-2) = 24/(-2) = -12

x₂ = (8 - 16)/(-2) = (-8)/(-2) = 4