Ответ(ы) на вопрос:
Гость1.
a) x/(2x+3)=1/x
x^2=2x+3x^2-2x-3=0D = 4 + 12 = 16X1=3X2=-2б) (2x+5)/(x^2+x)-2/x=3x/(x+1)(2x+5)/x(x+1) -2/x-3x/(x+1)=0(2x+5-2(x+1)-3x^2)/x(x+1) =0
(2x+5-2x-2-3x^2)/x(x+1) =0(3-3x^2)/x(x+1) =0(3-3x^2)/x(x+1) =0
3-3x^2=0 x(x+1)≠03(1-x^2)=0 x ≠0 x+1≠01-x^2=0 x≠-1(1-x)(1+x)=01-x=0 1+x=0X=1 x=-1 – п.к.
2.
а) 4x^4 - 17x^2 + 4 = 0
x^2 = t
4t^2 - 17t + 4 = 0
D = 289 - 64 = 225
t1 = 4 t2 = 0.25
x^2 = 4 x^2 = 0.25 x1 = 2 x3 = 0.5
x2 = -2 x4 = -0.5
б) (x^2 - 2x)^2 + (x^2 - 2x) = 12
x^2 - 2x = t
t^2 + t - 12 = 0
D = 1 + 48 = 49
t1 = 3 t2 = -4
x^2 - 2x = 3 x^2 - 2x = -4
x^2 - 2x - 3 = 0 x^2 - 2x + 4 = 0
D = 4 + 12 = 16 D = 4 - 16 = -12
x1 = 3
x2 = -1
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