Question

помогите пожалуйста!!!! если лень решать, натолкните хотя бы на мысль ​

Answer 1

√(x - 12/x) + 2√(x/(x² - 12)) = 3

одз не хочу писать

√((x² - 12)/x) + 2√(x/(x² - 12)) = 3

√((x² - 12)/x)  = y > 0

y + 2/y - 3 = 0

y² - 3y + 2 = 0

1. y = 1

√((x² - 12)/x)  = 1

(x² - 12) = x

x² - x - 12 = 0

D = 1 + 48 = 49

x12 = (1 +- 7)/2 = -3   4

x1 = -3

x2 = 4

оба корня , (x² - 12)/x > 0

2. y = 2

√((x² - 12)/x)  = 2

(x² - 12) = 4x

x² - 4x - 12 = 0

D = 16 + 48 = 64

x12 = (4 +- 8)/2 = -2   6

x1 = -2

x2 = 6

оба корня

Сумма -2 + 6 - 3 + 4 = 5

Answer 2

$\sqrt{x-\frac{12}{x} }+2\sqrt{\frac{x}{x^{2} -12} }=3\\\\\sqrt{\frac{x^{2}-12 }{x} }+2\sqrt{\frac{x}{x^{2} -12} }-3=0\\\\\sqrt{\frac{x^{2}-12 }{x} }=t \ \Rightarrow \ \sqrt{\frac{x}{x^{2} -12}}=\frac{1}{t} \ ; \ t>0\\\\t+\frac{2}{t} -3=0\\\\t^{2}-3t+2=0\\\\t_{1} =1\\\\t_{2}=2-teorema \ Vieta\\\\1)\sqrt{\frac{x^{2}-12 }{x} }=1\\\\(\sqrt{\frac{x^{2}-12 }{x} })^{2} =1^{2}\\\\\frac{x^{2}-12 }{x} =1\\\\x^{2}-x-12=0\\\\x_{1}=4\\\\x_{2}=-3$

$2)\sqrt{\frac{x^{2}-12 }{x} }=2\\\\(\sqrt{\frac{x^{2}-12 }{x} })^{2} =2^{2}\\\\\frac{x^{2}-12 }{x}=4\\\\x^{2}-4x-12=0\\\\x_{3}=6\\\\x_{4} =-2\\\\x_{1}+x_{2}+x_{3}+x_{4}= 4-3+6-2=\boxed5$