Giải phương trình sau : a) x ^ 2 – 5x + 5 = x – 3 b) 2x ^ 2 – 5x + 7 = x (x+3) -8

2 bình luận về “Giải phương trình sau : a) x ^ 2 – 5x + 5 = x – 3 b) 2x ^ 2 – 5x + 7 = x (x+3) -8”

1. a) x^2 – 5x + 5 = x – 3

   ⇔ x^2 – 5x – x + 5 + 3 = 0

   ⇔ x^2 – 6x  + 8 = 0

   ⇔ (x – 3)^2 – 1 = 0

   ⇔ (x – 3)^2 = 1

   ⇔ \(\left[ \begin{array}{l}x – 3=1\\x – 3=-1\end{array} \right.\) 

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

   Vậy S = {4 ; 2}

   b) 2x^2 – 5x + 7 = x(x + 3) – 8

   ⇔ 2x^2 – 5x + 7 = x^2 + 3x – 8

   ⇔ 2x^2 – x^2 – 5x – 3x + 7 + 8 = 0

   ⇔ x^2 – 8x + 15 = 0

   ⇔ (x – 4) = 1

   ⇔ \(\left[ \begin{array}{l}x – 4=1\\x – 4=-1\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x =5\\x = 3\end{array} \right.\) 

   Vậy S = {5 ; 3}

   Trả lời
2. a) x^2 – 5x +5 = x-3

   <=> x^2 – 6x + 8=0

   <=> x^2 – 4x – 2x + 8=0

   <=> (x-4)(x-2) =0

   <=> $\left[\begin{matrix} x=4\\ x=2\end{matrix}\right.$

   Vậy x=4 hoặc x=2

   b) 2x^2 – 5x +7 = x(x+3) – 8

   <=> 2x^2 – 5x + 7 = x^2 +3x – 8

   <=> x^2 – 8x + 15= 0

   <=> x^2 – 5x – 3x + 15 = 0

   <=> (x-5)(x-3) = 0

   <=>\(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\) 

   Vậy x = 5 hoặc x=3

   Trả lời