|(x-y) ² +2(xy+y ²-4y)| = xy+y ²-4y

1 bình luận về “|(x-y) ² +2(xy+y ²-4y)| = xy+y ²-4y”

1. |(x-y)^2 +2(xy+y^2 -4y)|=xy+y^2 -4y

   Do VT=|(x-y)^2 +2(xy+y^2 -4y)|^2 >=0=>xy+y^2 -4y>=0

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

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

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

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

   <=> $\begin{cases}x-y=0\\xy+y^2 -4y=0\end{cases}$

   <=> $\begin{cases}x=y\\y^2 +y^2 -4y=0\end{cases}$

   <=> $\begin{cases}x=y\\2y(y-2)=0\end{cases}$

   <=> $\begin{cases}x=y\\\left[ \begin{array}{l}y=0\\y=2\end{array} \right.\end{cases}$

   <=> \(\left[ \begin{array}{l}\begin{cases}x=0\\y=0\end{cases}\\\begin{cases}x=2\\y=2\end{cases}\end{array} \right.\) 

   Vậy (x;y)\in {(0;0),(2;2)} 

    

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