Question

[Thanh]

Cho \(a=xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)},b=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)

Hãy tính b theo a, biết x,y > 0

Answer 1

[Doraemon]

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

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

\(\Rightarrow b^2=a^2-1\)

Nếu \(x>0,y>0\Rightarrow b>0\Rightarrow b=\sqrt{a^2-1}\)

Nếu \(x< 0,\)\(y< 0\)\(\Rightarrow b< 0\Rightarrow b=-\sqrt{a^2-1}\)