Timf x a)x^2+3x=0 b)x^3-4x=0 c)x^2+5x=6

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1. a) $x^{2}$+3x=0

   ⇔ x(x + 3) = 0

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

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

   b) $x^{3}$ – 4x = 0

   ⇔ x($x^{2}$ – 4) = 0

   ⇔ \(\left[ \begin{array}{l}x=0\\x^{2}-4=0\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x=0\\x^2=4\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x=0\\x=±2\end{array} \right.\)

   c) $x^{2}$+5x=6

   ⇔ $x^{2}$ + 5x – 6 = 0

   ⇔ $x^{2}$ – x + 6x – 6 = 0 

   ⇔ x(x – 1) + 6(x – 1) = 0

   ⇔ (x – 1)(x + 6) = 0

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

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

    

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2. Đáp án:

    

   Giải thích các bước giải:

    a) `x^2+3x=0`

   `⇔ 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 `S={0;-3}`

   b) `x^3-4x=0`

   `⇔ x(x^2-4)=0`

   `⇔` \(\left[ \begin{array}{l}x=0\\x^2-4=0\end{array} \right.\)

   `⇔` \(\left[ \begin{array}{l}x=0\\x=±2\end{array} \right.\)  

   Vậy `S={0;-2;2}`

   c) `x^2+5x=6`

   `⇔ x^2+5x-6=0`

   `⇔ (x-1)(x+6)=0`

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

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

   Vậy `S={1;-6}`

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