Question

Tính S\(=x\sqrt{1+y^2}+y\sqrt{1+x^2}\) .Biết: \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)

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Answer 1

\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019^2\)

\(\Leftrightarrow M=2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019^2-2x^2y^2-x^2-y^2-1\)(Đặt M = .... cho gọn)

Có \(S=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)

\(\Rightarrow S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+M\)

\(\Rightarrow S^2=2x^2y^2+x^2+y^2+2019^2-2x^2y^2-x^2-y^2-1\)

\(\Rightarrow S=\sqrt{2019^2-1}\)