Question

xy +\(\sqrt{\left(x^2+1\right)\left(y^2+1\right)}\) =\(\sqrt{2017}\) . Tính giá trị của BT : A=\(x\sqrt{y^2+1}+y\sqrt{x^2+1}\)

Answer 1

Lời giải:

Ta có:

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

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

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

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

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

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

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