Question

[Sakura]

1/ tìm x,y nguyên dương thỏa mãn: \(x^2-y^2+2x-4y-10=0\)

2/ gtp nghiệm nguyên :\(x^2+2y^2+3xy+3x+5y=15\)

3/gtp nghiệm nguyên: \(2x^2+3xy-2y^2=7\)

Answer 1

[Mashiro Shiina]

1)\(x^2-y^2+2x-4y-10=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)-7=0\)

\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Leftrightarrow\left(x+1-y-2\right)\left(x+1+y+2\right)=7\)

\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)

Xét ước

2) \(x^2+2y^2+3xy+3x+3y=15\)

\(\Leftrightarrow x^2+y^2+2xy+y^2+xy+3x+3y=15\)

\(\Leftrightarrow\left(x+y\right)^2+y\left(x+y\right)+3\left(x+y\right)=15\)

\(\Leftrightarrow\left(x+y\right)\left(x+2y+3\right)=15\)

Xét ước

\(2x^2+3xy-2y^2=7\)

\(\Leftrightarrow2x^2+4xy-2y^2-xy=7\)

\(\Leftrightarrow2x\left(x+2y\right)-y\left(x+2y\right)=7\)

\(\Leftrightarrow\left(2x-y\right)\left(x+2y\right)=7\)

Xét ước