tim x: 6x(x-2012)+x-2012=0 x mu 2 +8x=0 x mu 2 -7x=0

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1.     6x(x-2012)+x-2012=0

   ⇔ (6x+1)(x-2012) = 0

   ⇔ \(\left[ \begin{array}{l}6x+1=0\\x-2012=0\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=-1/6\\x=2012\end{array} \right.\) 

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   x² +8x=0

   ⇔ x(x+8)=0

   ⇔ \(\left[ \begin{array}{l}x=0\\x+8=0\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=0\\x=-8\end{array} \right.\) 

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   x² -7x=0

   ⇔ x(x-7)=0

   ⇔ \(\left[ \begin{array}{l}x=0\\x-7=0\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=0\\x=7\end{array} \right.\) 

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2. Đáp án:

    

   Giải thích các bước giải:

   a) `6x(x-2012)+x-2012=0`

   `⇔ (6x+1)(x-2012)=0`

   `⇔` \(\left[ \begin{array}{l}6x+1=0\\x-2012=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=-\dfrac{1}{6}\\x=2012\end{array} \right.\) 

   Vậy `S={-\frac{1}{6};2012}` 

   b) `x^2+8x=0`

   `⇔ x(x+8)=0`

   `⇔` \(\left[ \begin{array}{l}x=0\\x+8=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=0\\x=-8\end{array} \right.\) 

   Vậy `S={0;-8}`

   c) `x^2-7x=0`

   `⇔ x(x-7)=0`

   `⇔` \(\left[ \begin{array}{l}x=0\\x-7=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=0\\x=7\end{array} \right.\) 

   Vậy `S={0;7}`

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