Question

tìm (x,y) nguyên thỏa mãn

(x-3).(x-5)+/y-2/=0

(x+1)(3-x)=2/y/+1

Answer 1

a: (x-3)(x-5)+|y-2|=0

=>(x-3)(x-5)=0 và y-2=0

hay \(\left(x,y\right)\in\left\{\left(3;2\right);\left(5;2\right)\right\}\)

b: \(\left(x+1\right)\left(3-x\right)=2\left|y\right|+1\)

\(\Leftrightarrow-x^2+3x+3-x-2\left|y\right|-1=0\)

\(\Leftrightarrow-x^2+2x+2-2\left|y\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x-2=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=3\\y=0\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\left\{\left(\sqrt{3}+1;0\right)\left(-\sqrt{3}+1;0\right)\right\}\)