Question

[trần tuấn phát]

tìm ngiệm nguyên dương 
\(\sqrt{x}+\sqrt{y}=\)\(\sqrt{50}\)

Answer 1

[Quỳnh Nguyễn]

\(\sqrt{50}=5\sqrt{2}\) \(\Leftrightarrow\sqrt{x}+\sqrt{y}=5\sqrt{2}\left(x,y\in Z^+\right)\)

Ta có: \(5\sqrt{2}=\sqrt{0}+5\sqrt{2}=\sqrt{2}+4\sqrt{2}=2\sqrt{2}+3\sqrt{2}\)

                         \(=5\sqrt{2}+\sqrt{0}=4\sqrt{2}+\sqrt{2}=3\sqrt{2}+2\sqrt{2}\)              

• \(\sqrt{x}+\sqrt{y}=\sqrt{0}+5\sqrt{2}=\sqrt{0}+\sqrt{50}\Rightarrow x=0;y=50\left(KTMDK\right)\)   
• \(\sqrt{x}+\sqrt{y}=\sqrt{2}+4\sqrt{2}=\sqrt{2}+\sqrt{32}\Rightarrow x=2;y=32\left(TMDK\right)\)
•  \(\sqrt{x}+\sqrt{y}=2\sqrt{2}+3\sqrt{2}=\sqrt{8}+\sqrt{18}\Rightarrow x=8;y=18\left(TMDK\right)\)  
• \(\sqrt{x}+\sqrt{y}=5\sqrt{2}+\sqrt{0}=\sqrt{50}+\sqrt{0}\Rightarrow x=50;y=0\left(KTMDK\right)\)     
• \(\sqrt{x}+\sqrt{y}=4\sqrt{2}+\sqrt{2}=\sqrt{32}+\sqrt{2}\Rightarrow x=32;y=2\left(TMDK\right)\)
• \(\sqrt{x}+\sqrt{y}=3\sqrt{2}+2\sqrt{2}=\sqrt{18}+\sqrt{8}\Rightarrow x=18;y=8\left(TMDK\right)\)

Vậy nghiệm của phương trình (x;y) = (2;32), (8;18), (32;2), (18;8)