b) (x – 1)2016+ (2y – 1)2016+ |x + 2y – z|2017= 0

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1. Đáp án:

   `b)(x-1)^2016+(2y-1)^2016+|x+2y-z|^2017=0`

   Vì `(x-1)^2016>=0`

   `(2y-1)^2016>=0`

   `|x+2y-z|^2017>=0`

   `=>(x-1)^2016+(2y-1)^2016+|x+2y-z|^2017>=0`

   `\text{Mà đề bài cho:}(x-1)^2016+(2y-1)^2016+|x+2y-z|^2017=0`

   `<=>` \(\begin{cases}x-1=0\\2x-1=0\\x+2y-z=0\\\end{cases}\)

   `<=>` \(\begin{cases}x=1\\2y=1\\z=x+2y=1+1\\\end{cases}\)

   `<=>` \(\begin{cases}x=1\\y=\dfrac12\\z=2\\\end{cases}\)

   Vậy `(x,y,z)=(1,1/2,2).`

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2. Giải:

   `(x – 1)^2016 >= 0` ; `(2y – 1)^2016 >= 0` ; `|x + 2y – z|^2017 >= 0` `(∀x,y,z)`

   Mà `(x – 1)^2016 + (2y – 1)^2016 + |x + 2y – z|^2017 = 0`

   `->` $\left\{ \begin{array}{l}(x – 1)^(2016) = 0\\(2y – 1)^(2016) = 0\\|x + 2y – z|^(2017) = 0\end{array} \right.$

   `->` $\left\{ \begin{array}{l}x – 1 = 0\\2y – 1 = 0\\x + 2y – z = 0\end{array} \right.$

   `->` $\left\{ \begin{array}{l}x = 1\\y = \dfrac{1}{2}\\1 + 1 – z = 0\end{array} \right.$

   `->` $\left\{ \begin{array}{l}x = 1\\y = \dfrac{1}{2}\\z = 2\end{array} \right.$

   Vậy `x = 1, y = 1/2, z = 2`

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