№4.
(x² - 3)/x + x/(x² -3) = 2 1/2
знаменатели не должны быть равны 0 :
x ≠0 ;
х²-3 ≠ 0 ⇒ х² ≠ 3 ; х≠√3 ; х≠ -√3
замена переменных :
(х² - 3)/x = t
x/(x² - 3) = 1/t
t + 1/t = 2 1/2
t≠0
(t² + 1) / t = 5/2
2(t²+1) = 5t
2t² + 2 - 5t = 0
2t² - 5t + 2 = 0
D = (-5)² - 4*2*2 = 25 - 16 = 9=3² D>0 два корня уравнения
t₁ = (5 -3)/ (2*2) = 2/4 = 1/2
t₂ = (5+3)/4 = 8/4 = 2
(x² - 3) / x = 1/2
2(x² - 3) = 1*x
2x² - 6 - x= 0
2x² - x - 6=0
D = (-1)² - 4*2*(-6) = 1 + 48 = 49=7² D>0
x₁ = (1-7)/(2*2) = -6/4 = -3/2 = -1.5
x₂ = (1+7)/4 = 8/4 = 2
(x² - 3) / x = 2
x² - 3 = 2x
x² - 2x - 3 = 0
D= (-2)² - 4*1*(-3) = 4 +12 = 16 = 4² D>0
x₃ = (2-4)/(2*1)= -1
x₄ = (2+4)/2 = 3
Ответ: х₁= -1,5 ; х₂=2 ; х₃ = -1 ; х₄ = 3
№5.
(х²+8)/х - 12х/(х² + 8) = 4
(x² +8)/x - (12/1) * (x/(x²+8)) = 4
(х² +8)/х = t
x/(x²+8) = 1/t
t - 12 * (1/t) = 4
t - 12/t = 4
(t² - 12)/t = 4
t² - 12 = 4t
t² - 4t - 12 = 0
D = (-4)² - 4*1*(-12) = 16 + 48 = 64=8² D>0
t₁ = (4-8)/(2*1) = -4/2=-2
t₂ = (4+8)/2 = 12/2 = 6
(x²+8)/x = -2
x² + 8 = -2x
x² + 2x + 8 = 0
D= 2² - 4*1*8 = 4 - 32 = - 28 D<0 вещественных корней уравнения нет<br>
(х² +8)/х = 6
х² +8 = 6х
х² - 6х + 8 = 0
D = (-6)² - 4*1*8 = 36-32 = 4 = 2² D>0 два корня уравнения
х₁ = (6 - 2) / (2*1) = 4/2 = 2
х₂ = (6+2)/2 = 4
ответ: х₁ = 2 ; х₂ = 4