Известно, что х+у/х-у + х-у/х+у=10 . Найдите значение выражения х^2/у^2+х^2-у^2 ОООЧень...

Замена (x+y)/(x-y) = t, тогда (x-y)/(x+y) = 1/t
t + 1/t = 10
t^2 - 10t + 1 = 0
D = 10^2 - 4*1*1 = 100 - 4 = 96 = (4√6)^2

1) t1 = (x+y)/(x-y) = (10 - 4√6)/2 = 5 - 2√6
x + y = (5 - 2√6)*x - (5 - 2√6)*y
(1 + 5 - 2√6)*y = (5 - 2√6 - 1)*x
y/x = (4 - 2√6)/(6 - 2√6) = (4 - 2√6)(6 + 2√6)/(36 - 4*6) = -4√6/12 = -√6/3
x^2/y^2 = 9/6 = 3/2
y/x = -√6/3
y^2 = 2/3*x^2
Из 1 уравнения
$\frac{(x+y)^2+(x-y)^2}{(x+y)(x-y)} = \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} = \frac{2x^2+2y^2}{x^2-y^2}=10 [/[tex] \frac{x^2+y^2}{x^2-y^2} =5$ tex]
x^2 - y^2 = (x^2 + y^2)/5 = (x^2 + 2/3*x^2)/5 = 5/3*x^2/5 = x^2/3
x^2/y^2 + x^2 - y^2 = 3/2 + x^2/3

2) t2 = (10 + 4√6)/2 = 5 + 2√6
x + y = (5 + 2√6)*x - (5 + 2√6)*y
(5 + 2√6+ 1)*y = (5 + 2√6 - 1)*x
(6 + 2√6)*y = (4 + 2√6)*x
y/x = (4 + 2√6)/(6 + 2√6) = (4 + 2√6)(6 - 2√6)/(36-4*6) = 4√6/12 = √6/3
x^2/y^2 = 9/6 = 3/2
y^2=2/3*x^2
Из 1 уравнения
$\frac{(x+y)^2+(x-y)^2}{(x+y)(x-y)} = \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} = \frac{2x^2+2y^2}{x^2-y^2}=10 [/[tex] \frac{x^2+y^2}{x^2-y^2} =5$ tex]
x^2 - y^2 = (x^2 + y^2)/5 = (x^2 + 2/3*x^2)/5 = 5/3*x^2/5 = x^2/3
x^2/y^2 + x^2 - y^2 = 3/2 + x^2/3

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