–12(x – 5) + 7(3 – x) = -4 (x – 2).(x + 4) = 0 (x –2).( x + 15) = 0 (7 – x).( x + 19)

0 bình luận về “–12(x – 5) + 7(3 – x) = -4 (x – 2).(x + 4) = 0 (x –2).( x + 15) = 0 (7 – x).( x + 19)”

1. `a)-12(x-5)+7(3-x)=-4`

   `→-12x+60+21-7x=-4`

   `→-19x+81=-4`

   `→-19x=-85`

   `→x=85/19`

   Vậy `x=85/19`

   `b)(x-2)(x+4)=0`

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

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

   Vậy `x∈{2;-4}`

   `c)(x-2)(x+15)=0`

   `⇔` \(\left[ \begin{array}{l}x-2=0\\x+15=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=2\\x=-15\end{array} \right.\) 

    Vậy `x∈{2;-15}`

   `d)(7-x)(x+19)=0`

   `⇔` \(\left[ \begin{array}{l}7-x=0\\x+19=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=7\\x=-19\end{array} \right.\) 

   Vậy `x∈{7;-19}`

    

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2. `a)  -12(x-5)+7(3-x)=-4`

   `<=> -12x+60+21-7x=-4`

   `<=> -12x-7x=-4-60-21`

   `<=> -19x=-85`

   `<=> x=85/19`

    

   `b)  (x-2)(x+4)=0`

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

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

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

   – Vậy `x in {2;-4}`

    

   `c)  (x-2)(x+15)=0`

   \(⇔\left[ \begin{array}{l}x-2=0\\x+15=0\end{array} \right.\)

   \(⇔\left[ \begin{array}{l}x=0+2\\x=0-15\end{array} \right.\)

   \(⇔\left[ \begin{array}{l}x=2\\x=-15\end{array} \right.\)

   – Vậy `x in {2;-15}`

    

   `d)  (7-x)(x+19)=0`

   \(⇔\left[ \begin{array}{l}7-x=0\\x+19=0\end{array} \right.\)

   \(⇔\left[ \begin{array}{l}-x=0-7\\x=0-19\end{array} \right.\)

   \(⇔\left[ \begin{array}{l}-x=-7\\x=-19\end{array} \right.\)

   \(⇔\left[ \begin{array}{l}x=7\\x=-19\end{array} \right.\)

   – Vậy `x in {7;-19}`

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