Question

1. Dwie liczby a i b ( a≠0, b≠0, a≠b ) spełniają warunek
\frac{a}{b} +a= \frac{b}{a} +b

Oblicz \frac{1}{a}+\frac{1}{b}

2. Sprowadź do najprostszej postaci wyrażenie:
\frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}

Answer 1

1.
(a² - 6b²)/ab = -1
a/b - 6b/a = -1
a/b = n
n - 6/n = -1
n^2 + n - 6 = 0
D = 25
n = - 3
n  =  2
(a² - 4b²)/2ab = a/2b - 2b/a = n/2 - 2/n
n = - 3
n/2 - 2/n = -3/2 + 2/3 = -5/6
n =  2
n/2 - 2/n = 2/2 - 2/2 = 0
(a² - 4b²)/2ab = -5/6
lub
(a² - 4b²)/2ab =  0
2.
a/(a-b)(a-c) + b/(b-c)(b-a) + c/(c-a)(c-b) =
=a/(a-b)(a-c) - b/(b-c)(a-b) + c/(a-c)(b-c)=
=a(b-c)/(a-b)(a-c)(b-c) - b(a-c)/(b-c)(a-c)(a-b) + c(a-b)/(a-c)(b-c)(a-b) =
=[a(b-c)- b(a-c)+ c(a-b)]/(a-b)(a-c)(b-c) =
= [ab - ac - ab + bc + ac - bc]/(a-b)(a-c)(b-c) =
= 0/(a-b)(a-c)(b-c) = 0