Tim x thuoc Z a) 210-5.(x-1) ²=-290 b) 41-2.|23-x|=37

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1. a, 210-5.(x-1) ²=-290

   ⇒ 5. (x-1)² = 210 – (-290) = 210 + 290

   ⇒ 5.(x-1)² = 500

   ⇒ (x-1)² = 100

   ⇒ \(\left[ \begin{array}{l}x-1=10\\x-1=-10\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}x=11\\x=-9\end{array} \right.\) 

   Vậy x ∈ {11 ; -9}

   b, 41-2.|23-x|=37

   ⇒ 2.|23 – x| = 41 – 37

   ⇒ 2.|23 – x| = 4

   ⇒ |23 – x| = 2

   ⇒ \(\left[ \begin{array}{l}23-x=2\\23-x=-2\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}x=21\\x=25\end{array} \right.\) 

   Vậy x ∈ {21 ; 25}

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2. a)

   `210-5(x-1)^2=-290`

   `⇒-5(x-1)^2=-290-210`

   `⇒-5(x-1)^2=-500`

   `⇒(x-1)^2=-500:(-5)`

   `⇒(x-1)^2=100`

   `⇒`\(\left[ \begin{array}{l}x-1=10\\x-1=-10\end{array} \right.\) 

   `⇒`\(\left[ \begin{array}{l}x=11\\x=-9\end{array} \right.\) 

    b)

   `41-2|23-x|=37`

   `⇒-2|23-x|=37-41`

   `⇒-2|23-x|=-4`

   `⇒|23-x|=-4:(-2)`

   `⇒|23-x|=2`

   `⇒`\(\left[ \begin{array}{l}23-x=2\\23-x=-2\end{array} \right.\) 

   `⇒`\(\left[ \begin{array}{l}x=21\\x=25\end{array} \right.\) 

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