( 4x ² – 25 ) ² – 9 ( 2x – 5 ) ² = 0 tim x

2 bình luận về “( 4x ² – 25 ) ² – 9 ( 2x – 5 ) ² = 0 tim x”

1. ( 4x^2 – 25 )^2 – 9 ( 2x – 5 )^2 = 0

   ⇔ [ ( 2x )^2 – 5^2 ]^2 – 9 ( 2x – 5 )^2 = 0

   ⇔ [ ( 2x + 5 ) . ( 2x – 5 ) ]^2 – 9 ( 2x – 5 ) ( 2x – 5 ) = 0

   ⇔ ( 2x + 5 )^2 . ( 2x – 5 )^2 – 9 ( 2x – 5 ) ( 2x – 5 ) = 0

   ⇔ ( 2x – 5 )^2 . [ ( 2x + 5 )^2 – 3^2 ] = 0

   ⇔ ( 2x – 5 )^2 . ( 2x + 5 + 3 ) . ( 2x + 5 – 3 ) = 0

   TH1:

   ( 2x – 5 )^2 = 0

   ⇔ 2x – 5 = 0

   ⇔ 2x = 5

   ⇔ x = 5/2

   TH2:

   2x + 5 + 3 = 0

   ⇔ 2x + 8 = 0

   ⇔ 2x = -8

   ⇔ x = -4

   TH3:

   2x + 5 – 3 = 0

   ⇔ 2x + 2 = 0

   ⇔ 2x = -2

   ⇔ x = -1

   Vậy x ∈ { 5/2 ; -4 ; -1 }

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

    (4x^2 – 25)^2-9(2x-5)^2=0

   ⇔ [(2x)^2 -5^2]^2 -9(2x-5)^2=0

   ⇔ [(2x-5)(2x+5)]^2 – 9(2x-5)^2=0

   ⇔ (2x-5)^2 (2x+5)^2 – 9(2x-5)^2=0

   ⇔ (2x-5)^2 [(2x+5)^2 – 9]=0

   ⇔ (2x-5)^2 (2x+5-3)(2x+5+3)=0

   ⇔ (2x-5)^2 (2x+2)(2x+8)=0

   ⇔ $\left[\begin{matrix} (2x-5)^2=0\\ 2x+2=0\\2x+8=0\end{matrix}\right.$

   ⇔ $\left[\begin{matrix} 2x-5=0\\ 2x=-2\\2x=-8\end{matrix}\right.$

   ⇔ $\left[\begin{matrix} x=\dfrac{5}{2}\\ x=-1\\x=-4\end{matrix}\right.$

   Vậy x in {5/2; -1;-4}

   Trả lời