Question

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

Trong đó xy > 0 . Tính b theo a .

 

Answer 1

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

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

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

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

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

\(=x^2+y^2+2x^2y^2+a^2-\left(x^2+y^2+2x^2y^2\right)-1=a^2-1\)\(\Leftrightarrow\)\(b=\sqrt{a^2-1}\) ( do a²>1 ) 

Cm: \(a^2>1\)

Có: \(1< \left(1+x^2\right)\left(1+y^2\right)\)\(\Leftrightarrow\)\(1< xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)\(\Leftrightarrow\)\(a^2>1\)