Question

Cho a = xy + \(\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\) và b = x\(\sqrt{1+y^2}\)+ y\(\sqrt{1+x^2}\). XY > 0. Tính b theo a

Answer 1

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

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

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

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

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

\(=a^2-1\Rightarrow b=\sqrt{a^2-1}\)