Question

Матеша 10 класс, если сделаете, то вкратце распишите Заранее благодарю.

Answer 1

6x² + 6/x² + 5x + 5/x - 38 = 0

6(x² + 1/x²) + 5(1/x + x) - 38 = 0

x ≠ 0

замена

1/x + x = t

(1/x + x)² = t²

1/x² + 2*1/x * x + x² = t²

1/x² + 2 + x² = t²

1/x² + x² = t² - 2

6(x² + 1/x²) + 5(1/x + x) - 38 = 0

6(t² - 2) + 5t - 38 = 0

6t² - 12 + 5t - 38 = 0

6t² + 5t - 50 = 0

D = 25 + 4*50*6 = 1225 = 35²

t12 = (-5 +- 35)/12 = 30/12 (5/2)   - 40/12 (-10/3)

обратно к х

1. 1/x + x = 5/2

2x² - 5x + 2 = 0

D = 25 - 16 = 9 = 3²

x12 = (5 +- 3)/4 = 2    1/2

2. 1/x + x = -10/3

3x² + 10x + 3 = 0

D = 100 - 36 = 64 = 8²

x12 = (-10 +- 8)/6 = -3  -1/3

ответ x = {2,1/2,-3,-1/3}

вкратце

Answer 2

$6x^{2} +\frac{6}{x^{2} } +5x+\frac{5}{x} -38 = 0$

$6(x^{2} +\frac{1}{x^{2} } )+5(x+\frac{1}{x}) -38 = 0$

Пусть  $x+ \frac{1}{x }$ = t , тогда :

$(x+ \frac{1}{x })^{2} =t^{2}$

$x^{2} +2+\frac{1}{x^{2} } =t^{2}$

$x^{2}+\frac{1}{x^{2} } =t^{2}-2$

----------------------------------

$6(t^{2}-2 )+5t -38 = 0$

$6t^{2}-12+5t -38 = 0$

$6t^{2}+5t - 50 = 0$

$D = 25+1200 = 1225=35^{2}$

$t_1 = \frac{-5+35}{12} =\frac{30}{12} =2.5$

$t_2 = \frac{-5-35}{12} =-\frac{40}{12} =- \frac{10}{3}$

Обратная замена :

$x+ \frac{1}{x }=2.5 | *2x$

$2x^{2} + 2=5x$

$2x^{2}-5x + 2=0$

$D = 25 - 16 = 9 = 3^{2}$

$x_1 = \frac{5+3}{4} =\frac{8}{4} =2$

$x_2 = \frac{5-3}{4} =\frac{2}{4} =0.5$

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$x+ \frac{1}{x }=-\frac{10}{3} | *3x$

$3x^{2} + 3=-10x$

$3x^{2}+10x + 3=0$

$D = 100 - 36 = 64=8^{2}$

$x_3 = \frac{-10+8}{6} =-\frac{2}{6} =-\frac{1}{3}$

$x_4 = \frac{-10-8}{6} =-\frac{18}{6} =-3$

ответ : $-3$ ,  $-\frac{1}{3}$ ,  $0.5$ ,  $2$