Question

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

b=\(x.\sqrt{1+y^2}+y.\sqrt{1+x^2}\) (x,y >0)

Tính b theo a.

Answer 1

Ta có:

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

\(\Rightarrow a^2=x^2+y^2+x^2y^2+1\)

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

\(\Rightarrow b^2=x^2+y^2+x^2y^2\)

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

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

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