Tìm x : 1, (2x + 1) . (x – 2) = 0 2, (x – 3) . (x + 4) = 0

0 bình luận về “Tìm x : 1, (2x + 1) . (x – 2) = 0 2, (x – 3) . (x + 4) = 0”

1. Đáp án:

   1,` (2x  +1) . (x – 2) = 0`

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

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

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

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

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

   Vậy `x = -0,5` hoặc `x = 2`

   2,` (x – 3) . (x + 4) = 0`

   `⇒` \(\left[ \begin{array}{l}x – 3 = 0\\x+4 = 0\end{array} \right.\) 

   `⇒` \(\left[ \begin{array}{l}x = 0 + 3\\x = 0-4\end{array} \right.\) 

   `⇒` \(\left[ \begin{array}{l}x = 3\\x = -4\end{array} \right.\) 

   Vậy `x = 3` hoặc `x = -4`

   Bình luận

2. 1, $(2x+1).(x-2)=0$

   $TH1: 2x+1=0$

   $2x=0-1$

   $2x=-1$

   $x=\dfrac{-1}{2}$

   $TH2: x-2=0$

   $x=0+2$

   $x=2$

   Vậy `x∈{\frac{-1}{2},2}`

   2, $(x-3).(x+4)=0$

   $TH1: x-3=0$

   $x=0+3$

   $x=3$

   $TH2: x+4=0$

   $x=0-4$

   $x=-4$

   Vậy `x∈{3,-4}`

   Bình luận