a) -4x – 9= x+11 b) 3x – 4= 5x+11 c) (2x-5)(6-x) = 0 d) 2x(x-3) – x+3= 0

0 bình luận về “a) -4x – 9= x+11 b) 3x – 4= 5x+11 c) (2x-5)(6-x) = 0 d) 2x(x-3) – x+3= 0”

1. a)    -4x – 9 = x +11

   <=>-4x – x = 11 + 9

   <=>-5x = 20

   <=>x = -4

   Vậy S ={-4}

   b)    3x – 4 = 5x + 11

   <=>3x – 5x = 11 + 4

   <=>-2x = 15

   <=>x = -7,5

   Vậy S = {-7,5}

   c)    (2x – 5)(6 – x) = 0

    TH1: 2x -5 =0                          TH2: 6 – x = 0

         <=>2x = 5                              <=>-x = -6

         <=>x = 5/2                             <=>x = 6

   Vậy S = {5/2 ; 6}

   d)    2x(x – 3) – x + 3 = 0

   <=>2x(x – 3) – (x – 3 = 0

   <=>(2x – 1)(x – 3) = 0

   TH1: 2x – 1 = 0                       TH2: x – 3 = 0

   <=>2x = 1                             <=> x = 3

   <=>x = 1/2

   Vậy S = {1/2 ; 3}

   Trả lời

2. Đáp án + Giải thích các bước giải:

    `a)` `-4x-9=x+11`

   `<=>-4x-x=11+9`

   `<=>-5x=20`

   `<=>x=-4`

   Vậy `S={-4}`

   `b)` `3x-4=5x+11`

   `<=>3x-5x=11+4`

   `<=>-2x=15`

   `<=>x=-15/2`

   Vậy `S={-15/2}`

   `c)` `(2x-5)(6-x)=0`

   `<=>` \(\left[ \begin{array}{l}2x-5=0\\6-x=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=5\\-x=-6\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=6\end{array} \right.\) 

   Vậy `S={6;5/2}`

   `d)` `2x(x-3)-x+3=0`

   `<=>2x(x-3)-(x-3)=0`

   `<=>(2x-1)(x-3)=0`

   `<=>` \(\left[ \begin{array}{l}2x-1=0\\x-3=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=1\\x=3\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=3\end{array} \right.\)

   Vậy `S={1/2;3}`

   Trả lời