Question

[Nguyen Thuy Linh]

Cho |3x-1|²⁰¹⁵ +(2x-y)²⁰¹⁶≤0

Tính A=-2x²-xy+y²+2016

Answer 1

[Mashiro Shiina]

\(\left|3x-1\right|^{2015}+\left(2x-y\right)^{2016}\le0\)

\(\left\{{}\begin{matrix}\left|3x-1\right|\ge0\Rightarrow\left|3x-1\right|^{2015}\ge0\forall x\\\left(2x-y\right)^{2016}\ge0\forall x;y\end{matrix}\right.\)

\(\Rightarrow\left|3x-1\right|^{2015}+\left(2x-y\right)^{2016}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|3x-1\right|^{2015}+\left(2x-y\right)^{2016}\ge0\\\left|3x-1\right|^{2015}+\left(2x-y\right)^{2016}\le0\end{matrix}\right.\)

\(\Rightarrow\left|3x-1\right|^{2015}+\left(2x-y\right)^{2016}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|3x-1\right|^{2015}=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\\\left(2x-y\right)^{2016}=0\Rightarrow2x=y\Rightarrow x=\dfrac{1}{2}y\Rightarrow y=\dfrac{1}{6}\end{matrix}\right.\)

\(\Rightarrow A=-2\dfrac{1}{3}^2-\dfrac{1}{3}.\dfrac{1}{6}+\dfrac{1}{6}^2+2016\)

\(A=-2.\dfrac{1}{9}-\dfrac{1}{18}+\dfrac{1}{36}+2016\)

\(A=\dfrac{-8}{36}-\dfrac{2}{36}+\dfrac{1}{36}+2016\)

\(A+-\dfrac{1}{4}+2016\)