Question

URGENT!!Numerele reale nenule a si b satisfac egalitatile $a+b= \frac{1}{a}+\frac{1}{b}=$6. Sa se afle valoarea numerica a expresiei $E= \frac{a}{b}+\frac{b}{a}+1979$.

Answer 1

1/a + 1/b = 6 <=> (a+b)/(ab) = 6 <=> 6/(ab) = 6 <=> ab = 1 ;
a/b + b/a = (a^2 + b^2)/(ab) = [(a+b)^2 - 2ab]/(ab) = (36 - 12)/1 = 24;
In final, 24 + 1979 = 2003.
Bafta!