Question

Answer 1

Ta có: \(3x^2-4xy+2y^2=3\)

=>\(2x^2-4xy+2y^2+x^2=3\)

=>\(2\left(x-y\right)^2+x^2=3\)

=>\(x^2<=3\)

mà x là số nguyên dương

nên x=1

Ta có: \(2\left(x-y\right)^2+x^2=3\)

=>\(2\left(1-y\right)^2=3-1^2=2\)

=>\(\left(y-1\right)^2=1\)

=>\(\left[\begin{array}{l}y-1=-1\\ y-1=1\end{array}\right.\Rightarrow\left[\begin{array}{l}y=0\left(loại\right)\\ y=2\left(nhận\right)\end{array}\right.\)

\(M=x^{2022}+\left(y-3\right)^{2022}\)

\(=1^{2022}+\left(2-3\right)^{2022}=1+1=2\)