Question

Помогите
Даю 20 баллов ​

Answer 1

$1) \frac{2x+3}{x+2} = \frac{3x+2}{x}\\\\x\neq-2; x\neq0\\\\(2x+3)(x) = (3x+2)((x+2)\\\\2x^2+3x = 3x^2+6x+2x+4\\\\2x^2 - 3x^2 + 3x - 6x - 2x - 4 = 0\\\\-x^2-5x - 4 = 0|:(-1)\\\\x^2 + 5x + 4 = 0\\\\D = 25 - 16 = 9\\\\\sqrt{D} = 3\\\\x_1=\frac{-5-3}{2} = -\frac{8}{2} = -4\\\\ x_2=\frac{-5+3}{2} = -\frac{2}{2} = -1$

$\frac{y+3}{y-3} = \frac{2y+3}{y}\\\\y\neq3; y\neq0\\\\(y+3)(y) = (2y+3)(y-3) \\\\ y^2+3y = 2y^2-6y+3y-9\\\\y^2-2y^2 +3y+6y -3y + 9 =0\\\\-y^2 + 6y +9 = 0\\\\y^2 - 6y - 9 = 0\\\\D = 36 + 36 = 72\\\\\sqrt{D} = 6\sqrt{2}\\\\y_1 = \frac{6-6\sqrt{2}}{2} = 3-3\sqrt{2}\\\\y_2 = \frac{6+6\sqrt{2}}{2} = 3+3\sqrt{2}$

Удачи!