Question

Cho \(x\times y+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2018\) . Tính \(T=x\sqrt{1+y^2}+y\sqrt{1+x^2}\) .

Answer 1

Lời giải:

Ta có:

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

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

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

\(\Leftrightarrow 2x^2y^2+x^2+y^2+2xy\sqrt{(x^2+1)(y^2+1)}=2018^2-1\)

\(\Leftrightarrow x^2(y^2+1)+y^2(x^2+1)+2xy\sqrt{(x^2+1)(y^2+1)}=2018^2-1\)

\(\Leftrightarrow (x\sqrt{y^2+1}+y\sqrt{x^2+1})^2=2018^2-1\)

\(\Leftrightarrow T^2=2018^2-1\)

Do đó: \(T=\pm \sqrt{2018^2-1}\)