Question

Как решить уравнение x^2+3x/x-2= 10/x-2

Answer 1

$\frac{ {x}^{2}}{1} + \frac{3x}{x - 2 } = \frac{10}{x - 2} \\ {x}^{3} - 2 {x}^{2} + 3x = 10 \\ x( {x}^{2} - 2x + 3) = 10 \\ x = 10 \\ {x}^{2} - 2x + 3 = 10 \\ {x}^{2} - 2x - 7 = 0 \\ d = 4 - 4 \times 1 \times - 7 = 4 + 28 = 32 = 5 \sqrt{2 } \\ x1 = \frac{2 - 5 \sqrt{2} }{2 \times 1} = \frac{2(1 - 2.5 \sqrt{2)} }{2} = 1 - 2.5 \sqrt{2} \\ x2 = \frac{2 + 5 \sqrt{2} }{2 \times 1} = 1 + 2.5 \sqrt{2}$

Answer 2

$\dfrac{x^2+3x}{x-2}=\dfrac{10}{x-2}\ \ \ ,\ \ \ \ \ ODZ: x\ne 2\\\\\\x^2+3x=10\\\\x^2+3x-10=0\\\\x_1=-5\ ,\ x_2=2\ \ \ \ (teorema\ Vieta)\\\\x_2=2\notin ODZ\\\\Otvet:\ \ x=-5\ .$