Đáp án: $A=\dfrac12$
Giải thích các bước giải:
Đặt $x=a^2, y=b^2, a,b>0\to \sqrt{xy}=ab$
Ta có:
$\dfrac{1}{1+x}+\dfrac{1}{1+y}=\dfrac{2}{1+\sqrt{xy}}$
$\to \dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}=\dfrac{2}{1+ab}$
$\to( \dfrac{1}{1+a^2}-\dfrac{1}{1+ab})+(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab})=0$
$\to \dfrac{1+ab-(1+a^2)}{(1+a^2)(1+ab)}+\dfrac{1+ab-(1+b^2)}{(1+b^2)(1+ab)}=0$
$\to \dfrac{ab-a^2}{(1+a^2)(1+ab)}+\dfrac{ab-b^2}{(1+b^2)(1+ab)}=0$
$\to \dfrac{a(b-a)}{(1+a^2)(1+ab)}+\dfrac{b(a-b)}{(1+b^2)(1+ab)}=0$
$\to \dfrac{a(b-a)}{(1+a^2)(1+ab)}-\dfrac{b(b-a)}{(1+b^2)(1+ab)}=0$
$\to \dfrac{b-a}{1+ab}(\dfrac{a}{1+a^2}-\dfrac{b}{1+b^2})=0$
$\to \dfrac{b-a}{1+ab}\cdot \dfrac{a(1+b^2)-b(1+a^2)}{(1+a^2)(1+b^2)}=0$
$\to \dfrac{b-a}{1+ab}\cdot \dfrac{(a-b)+ab(b-a)}{(1+a^2)(1+b^2)}=0$
$\to \dfrac{b-a}{1+ab}\cdot \dfrac{-(b-a)+ab(b-a)}{(1+a^2)(1+b^2)}=0$
$\to \dfrac{(b-a)^2}{1+ab}\cdot \dfrac{(ab-1)}{(1+a^2)(1+b^2)}=0$
$\to (b-a)^2(ab-1)=0$
Vì $x,y$ phân biệt $\to x\ne y\to a\ne b\to a-b\ne 0$
$\to ab-1=0$
$\to ab=1$
$\to \sqrt{xy}=1$
$\to \dfrac{1}{1+x}+\dfrac{1}{1+y}=\dfrac{2}{1+1}=1$
$\to A=1-\dfrac{1}{1+1}=\dfrac12$
