Có:\(\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)=5\)
=> \(\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)\left(y-\sqrt{y^2+5}\right)=5\left(y-\sqrt{y^2+5}\right)\)
<=> \(\left(x+\sqrt{x^2+5}\right)\left(y^2-y^2-5\right)=5\left(y-\sqrt{y^2+5}\right)\)
<=> \(-5\left(x+\sqrt{x^2+5}\right)=5\left(y-\sqrt{y^2+5}\right)\)
<=> \(-x-\sqrt{x^2+5}=y-\sqrt{y^2+5}\)
<=> \(-\sqrt{x^2+5}+\sqrt{y^2+5}=y+x\)(1)
Có :\(\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)=5\)
=> \(\left(x+\sqrt{x^2+5}\right)\left(x-\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)=5\left(x-\sqrt{x^2+5}\right)\)
<=> \(\left(x^2-x^2-5\right)\left(y+\sqrt{y^2+5}\right)=5\left(x-\sqrt{x^2+5}\right)\)
<=>\(-5\left(y+\sqrt{y^2+5}\right)=5\left(x-\sqrt{x^2+5}\right)\)
<=> \(-y-\sqrt{y^2+5}=x-\sqrt{x^2+5}\)
<=> \(-\sqrt{y^2+5}+\sqrt{x^2+5}=x+y\) (2)
Cộng vế vs vế của (1),(2) có
\(-\sqrt{x^2+5}+\sqrt{y^2+5}-\sqrt{y^2+5}+\sqrt{x^2+5}=y+x+x+y\)
<=> 0=2(x+y)
<=> 0=x+y=E
Vậy E=0
