Question

Cho x > 0; y > 0 thỏa mãn:

xy+ \(\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2009}\)

Tính:

A= x\(\sqrt{y^2+1}+y\sqrt{x^2+1}\)

Answer 1

\(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2009}\)

\(x^2y^2+\left(x^2+1\right)\left(y^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2009\)

\(x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2009\)

\(x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2008\)

\(\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2008\)

\(\Leftrightarrow A^2=2009\)

\(\Leftrightarrow A=\sqrt{2009}\) khi x, y > 0 hoặc \(A=-\sqrt{2009}\) khi x, y < 0