Question

Cho các số tự nhiên x,y thỏa mãn: \(2x^2+8y^2-4xy-4y-2x+2=0\)

TÍnh giá trị biểu thức: B = \(\dfrac{\left(x+2y-2\right)^{2025}-x^{2025}}{y}\)

Answer 1

Ta có: \(2x^2+8y^2-4xy-4y-2x+2=0\)

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

=>\(\left(x-2y\right)^2+\left(x-1\right)^2+\left(2y-1\right)^2=0\)

=>\(\left\{{}\begin{matrix}x-2y=0\\x-1=0\\2y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(B=\dfrac{\left(x+2y-2\right)^{2025}-x^{2025}}{y}\)

\(=\dfrac{\left(1+2\cdot\dfrac{1}{2}-2\right)^{2025}-1^{2025}}{\dfrac{1}{2}}=\dfrac{-1}{\dfrac{1}{2}}=-2\)