x^2(-3 + 2x)^2 – (4x^2 – 2x + 1)^2 = 0
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$\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
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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) = 0Mà (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}.