Question

Cho x,y >0 thỏa mãn: \(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2018}\)

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

Answer 1

\(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2018}\)

\(x^2y^2+\left(x^2+1\right)\left(y^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2018\)

\(x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2018\)

\(x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2017\)

\(\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2017\)

\(\Rightarrow A=\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2017\)

\(\Rightarrow A=\sqrt{2017}\) khi x, y > 0 hoặc \(A=-\sqrt{2017}\) khi x, y < 0