tìm x thuộc Z biết x(x+3)=0 (x-2)(5-x)=0 (x-1)(x^2+1)=0

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1. +)

   $x(x+3)=0$

   ⇒\(\left[ \begin{array}{l}x=0\\x+3=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\) 

   Vậy $x=\{0;-3\}$

   +)

   $(x-2)(5-x)=0$

   ⇒\(\left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\) 

   Vậy $x=\{2;5\}$

   +)

   $(x-1)(x²+1)=0$

   ⇒\(\left[ \begin{array}{l}x-1=0\\x^2+1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1\\x^2=-1(loại)\end{array} \right.\) 

   Vậy $x=1$

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2. Đáp án:

    

   Giải thích các bước giải:

   x(x+3)=0

   ⇒\(\left[ \begin{array}{l}x=0\\x+3=0\end{array} \right.\)

   ⇒\(\left[ \begin{array}{l}x=0\\x=0-3\end{array} \right.\)

   ⇒\(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\) 

    

   (x-2)(5-x)=0

   ⇒\(\left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right.\) 

   ⇒\(\left[ \begin{array}{l}x=0+2\\x=5-0\end{array} \right.\)

   ⇒\(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)

   (x-1)(x²+1)=0

   ⇒\(\left[ \begin{array}{l}x-1=0\\x^2+1=0\end{array} \right.\) 

   ⇒\(\left[ \begin{array}{l}x=0+1\\x=0-1\end{array} \right.\) 

   ⇒\(\left[ \begin{array}{l}x=-1\\x^2=-1\end{array} \right.\) 

   ⇒\(\left[ \begin{array}{l}x=-1\\x^2=-1^2\end{array} \right.\)

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

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