Ta có :
\(\sqrt{x}+\sqrt{y}=\sqrt{981}\)
\(\Leftrightarrow\left(\sqrt{x}+\sqrt{y}\right)^2=981\)
\(\Leftrightarrow x+y+2\sqrt{xy}=981\)
\(\Leftrightarrow x+y+2\sqrt{xy}-x+y=981-x+y\)
\(\Leftrightarrow\left(\sqrt{x}+\sqrt{y}\right)^2-\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)=981-x+y\)
\(\Leftrightarrow2\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)=981-x+y\)
\(\Leftrightarrow\left[2\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\right]^2=\left(981-x+y\right)^2\)
\(\Leftrightarrow\left(2\sqrt{y}.\sqrt{981}\right)^2=\left(981-x+y\right)^2\)
\(\Leftrightarrow4y.981=\left(981-x+y\right)^2\)
Vì \(981=3^2.109\Rightarrow2^2.3^2.y.109=\left(981-x+y\right)^2\)
Vế phải : \(\left(981-x+y\right)^2\)là số chính phương \(\Rightarrow y=k^2.109\)với k là số tự nhiên
Vậy phương trình có hai nghiệm : \(\left(x;y\right)=\left(0;981\right);\left(981;0\right)\)
