( x – 2 ) ( x – 3 ) – x^2 + 4 = 0 ( 2x – 3 )^2 – 16 = 0

2 bình luận về “( x – 2 ) ( x – 3 ) – x^2 + 4 = 0 ( 2x – 3 )^2 – 16 = 0”

1. (x-2)(x-3)-x^2 + 4=0

   ⇔ (x-2)(x-3) -(x^2 -4)=0

   ⇔ (x-2)(x-3)-(x-2)(x+2)=0

   ⇔ (x-2)[x-3-(x+2)]=0

   ⇔ (x-2)(x-3-x-2)=0

   ⇔ -5(x-2)=0

   ⇔ x-2=0

   ⇔ x=2

   Vậy x=2

   (2x-3)^2 -16=0

   ⇔ (2x-3)^2 – 4^2 =0

   ⇔ (2x-3 – 4)(2x-3 +4)=0

   ⇔ (2x-7)(2x+1)=0

   ⇔ \(\left[ \begin{array}{l}2x-7=0\\2x+1=0\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}2x=7\\2x=-1\end{array} \right.\) 

   ⇔ $\left[\begin{matrix} x= \dfrac{7}{2}\\ x= \dfrac{-1}{2}\end{matrix}\right.$

   Vậy x ∈ {7/2; (-1)/2}

   Trả lời
2. (x – 2) (x – 3) – x^2 + 4 = 0

   <=> x^2 – 2x – 3x + 6 – x^2 + 4 = 0

   <=> -2x – 3x + 6 + 4 = 0

   <=> -5x + 10 = 0

   <=> -5x = -10

   <=> x = (-10)/(-5)

   <=> x = 2

   Vậy S = {2}

                                                                       

   (2x – 3)^2 – 16 = 0

   <=> (2x – 3)^2 – 4^2 = 0

   <=> (2x – 3 – 4) (2x – 3 + 4) = 0

   <=>\(\left[ \begin{array}{l}2x-3-4=0\\2x-3+4=0\end{array} \right.\) 

   <=>\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{array} \right.\) 

   Vậy S = {7/2 ; – 1/2}

   Trả lời