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

2 bình luận về “x^2(-3 + 2x)^2 – (4x^2 – 2x + 1)^2 = 0”

1. $\text{→ Giải đáp + Lời giải và giải thích chi tiết:}$

   $\text{→ Ta có :}$

   $\text{x²( 2x – 3 )² – ( 4x² – 2x + 1 )² = 0}$

   $\text{⇔ [ x( 2x – 3 ) + 4x² – 2x + 1 ][ x( 2x – 3 ) – ( 4x² – 2x + 1 ) ] = 0}$

   $\text{⇔ ( 2x² – 3x + 4x² – 2x + 1 )( 2x² – 3x – 4x² + 2x – 1 ) = 0}$

   $\text{⇔ ( 6x² – 5x + 1 )( – 2x²- x – 1 ) = 0}$

   $\text{⇔ ( 6x² – 5x + 1 )( 2x² + x + 1 ) = 0}$

   $\text{→ Xét 2x² + x + 1 = ( 2x² + x + $\dfrac{1}{8}$ ) + $\dfrac{7}{8}$}$

   $\text{= ( $\sqrt{2}$x + $\dfrac{1}{\sqrt{8}}$ )² + $\dfrac{7}{8}$ > 0  ;( $\forall$ x )}$

   $\text{⇒ 6x² – 5x + 1 = 0 ⇔ 6x² – 3x – 2x + 1 = 0 ⇔ 3x( 2x – 1 ) – ( 2x – 1 ) = 0}$

   $\text{⇔ ( 2x – 1 )( 3x – 1 ) = 0}$

   $\text{⇔ $\left[\begin{matrix} x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.$}$

   $\text{→ Vậy S = { $\dfrac{1}{2}$ ; $\dfrac{1}{3}$ } }$

   5 sao nha 

   Trả lời
2. Giải đáp + Lời giải và giải thích chi tiết:

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

   <=> [x(2x – 3)]^2 – (4x^2 – 2x + 1)^2 = 0

   <=> (2x^2 – 3x)^2 – (4x^2 – 2x + 1)^2 = 0

   <=> (2x^2 – 3x – 4x^2 + 2x – 1)(2x^2 – 3x + 4x^2 – 2x + 1) = 0

   <=> (-2x^2 – x – 1)(6x^2 – 5x + 1) = 0

   <=> -(2x^2 + x + 1)(6x^2 – 2x – 3x + 1) = 0

   <=> 2(x^2 + 1/2x + 1/2)[2x(3x – 1) – (3x – 1)] = 0

   <=> (x^2 + 1/2x + 1/16 + 7/16)(2x – 1)(3x – 1) = 0

   <=> [(x + 1/4)^2 + 7/16](2x – 1)(3x – 1) = 0

   Mà (x + 1/4)^2 >= 0 AA x

   => (x + 1/4)^2 + 7/16 >= 7/16 > 0 AA x

   <=> (2x -1)(3x – 1) = 0

   <=> [(x = 1/2),(x = 1/3):}

   Vậy S = {1/2; 1/3}.

   Trả lời