Question

Cho x;y > 0; x,y thỏa mãn \(\left(x+1\right)\left(y+1\right)=2\). Tính:

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

Answer 1

\(\left(x+1\right)\left(y+1\right)=2\)

\(\Leftrightarrow x=\frac{1-y}{1+y}\)

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

\(=\sqrt{\left(\frac{1-y}{1+y}\right)^2+y^2-\sqrt{2\left(\left(\frac{1-y}{1+y}\right)^2+1\right)\left(y^2+1\right)}+2}+\left(\frac{1-y}{1+y}\right)y\)

\(=\sqrt{\left(\frac{1-y}{1+y}\right)^2+y^2-2.\frac{y^2+1}{y+1}+2}+\left(\frac{1-y}{1+y}\right)y\)

\(=\sqrt{\left(\frac{y^2+1}{y+1}\right)^2}+\left(\frac{1-y}{1+y}\right)y\)

\(=\frac{y^2+1}{y+1}+\left(\frac{1-y}{1+y}\right)y=1\)