Từ giả thiết ta có \(2016=x^2y^2+1+x^2+y^2+x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

\(\Leftrightarrow x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2015\)

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

\(=x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2015\)

\(\Rightarrow S=\sqrt{2015}\) (Vì S > 0)

2

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)

A= \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-2\right)^2}=\left|3x-1\right|+\left|3x-2\right|\)

ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1

=> A ≥ 1

=> Min A =1 khi 1/3 ≤ x ≤ 2/3

nhận liên hợp ta có  \(\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=x^2+1-x^2=1\)

mà theo đề bài ta có \(\left(\sqrt{x^2+1}+x\right)\left(y+\sqrt{y^2+1}\right)=1\)

==> \(\sqrt{x^2+1}-x=y+\sqrt{y^2+1}\)

tương tự ta có \(\sqrt{x^2+1}+x=\sqrt{y^2+1}-y\)

trừ từng vế 2 pt trên ta có 2x=-2y <=>x=-y

đến đây ok rùi nhé bạn 

\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=\sqrt{2000}\)

\(\Rightarrow x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2000\)

\(\Rightarrow2x^2y^2+x^2+y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=1999\)

Ta có:

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

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

\(\Rightarrow S^2=1999\Rightarrow S=\pm\sqrt{1999}\)

o my god

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

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

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

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

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

\(\Leftrightarrow\left[y\left(1+x^2\right)+x\left(1+y^2\right)\right]^2=2018\)

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

hay \(A=\sqrt{2018}\)

Lời giải:

Ta có:

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

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

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

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

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

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

\(\Leftrightarrow A^2=2017-1=2016\Rightarrow A=\sqrt{2016}\)