Question

Cho \(5x^2-4xy+y^2-4x+4=0\)

Tính A= \(\left(x-1\right)^3+\left(y+2\right)^2\)

Answer 1

\(5x^2-4xy+y^2-4x+4=0\)

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

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

Do \(\left(2x-y\right)^2,\left(x-2\right)^2\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

\(A=\left(x-1\right)^3+\left(y+2\right)^3=\left(2-1\right)^3+\left(4+2\right)^3\)

\(=1+6^3=217\)

Answer 2

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