giải phương trình a,(x+2)(3x+1)=(2x+4)(x+1) x ²_4+5 ×(4x_0)=0

1 bình luận về “giải phương trình a,(x+2)(3x+1)=(2x+4)(x+1) x ²_4+5 ×(4x_0)=0”

1. $\text{ a) (x+2)(3x + 1) = (2x + 4)(x+1)}$

   $\text{<-> (x+2)(3x + 1) = 2(x + 2)(x+1)}$

   $\text{<-> (x+2)(3x + 1) – 2(x + 2)(x+1) = 0}$

   $\text{<-> (x+2)[(3x + 1) – 2(x+1)] = 0}$

   $\text{<-> (x+2)(3x + 1 – 2x – 2) = 0}$

   $\text{<-> (x+2)(x – 1) = 0}$

   $\text{<-> x+2 = 0 hoặc x – 1 = 0}$

   $\text{<-> x = -2 hoặc x  = 1}$

   $\text{Vậy S = {-2; 1}}$

   $\text{ b) x^2 – 4 + 5 $\times$ (4x – 8) = 0}$

   $\text{<-> x^2 – 2^2 + 5 $\times$ (4x – 8) = 0}$

   $\text{<-> (x – 2)(x + 2) + 5 $\times$ 4(x – 2) = 0}$

   $\text{<-> (x – 2)(x + 2) + 20 $\times$ (x – 2) = 0}$

   $\text{<-> (x – 2)[(x + 2) + 20]= 0}$

   $\text{<-> (x – 2)(x + 2 + 20)= 0}$

   $\text{<-> (x – 2)(x + 22)= 0}$

   $\text{<-> x – 2= 0 hoặc x + 22 = 0}$

   $\text{<-> x = 2 hoặc x = -22}$

   $\text{Vậy S ={ 2; -22}}$

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