giải pt (x – 1)(2x² – 10) = 0 (2x – 7)² – 6(2x – 7)(x – 3) = 0 (5x + 3)(x² + 4) = 0

0 bình luận về “giải pt (x – 1)(2x² – 10) = 0 (2x – 7)² – 6(2x – 7)(x – 3) = 0 (5x + 3)(x² + 4) = 0”

1. a, (x – 1)(2x² – 10) = 0

   ⇔\(\left[ \begin{array}{l}x – 1=0\\2x² – 10=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x =1\\x² =5\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x =1\\x=+- √5\end{array} \right.\) 

   KL:..

   b, (2x – 7)² – 6(2x – 7)(x – 3) = 0

   ⇔(2x – 7).[2x – 7-6(x-3)]=0

   ⇔\(\left[ \begin{array}{l}2x – 7=0\\2x – 7-6x+18=0end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=7/2\\11=4x\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x =7/2\\x=11/4\end{array} \right.\) 

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2.     (x – 1)(2x² – 10) = 0

   ⇔ x – 1 = 0 hoặc 2x² – 10 = 0

   ⇔ x = 1 hoặc 2x² = 10

   ⇔ x = 1 hoặc x² = 5

   ⇔ x = 1 hoặc x = √5

   Vậy S = { 1 ; √5 }.

       (2x – 7)² – 6(2x – 7)(x – 3) = 0

   ⇔ (2x – 7)² = 6(2x – 7)(x – 3)

   ⇔ (2x – 7)² = (2x – 7)(6x – 18)

   ⇔ (2x – 7)² – (2x – 7)(6x – 18) = 0

   ⇔ (2x – 7)[(2x – 7) – (6x – 18)] = 0

   ⇔ (2x – 7)(2x – 7 – 6x + 18) = 0

   ⇔ (2x – 7)(-4x + 11) = 0
   ⇔ 2x -7 = 0 hoặc -4x + 11 = 0
   ⇔ 2x = 7 hoặc -4x = -11

   ⇔ x = 7/2 hoặc x = 11/4

   Vậy S = { 7/2 ; 11/4 }.

       (5x + 3)(x² + 4) = 0

   ⇔ 5x + 3 = 0 hoặc x² + 4 = 0

   ⇔ 5x = -3 hoặc x² = -4

   ⇔ x = -3/5 hoặc x = -√4

   Vậy S = { -3/5 ; -√4 }

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