Giải phương trình: ` 5/(x^2 +1) + 7/(x^2+3) – (6+3x^2)/(x^2+5)=0`

1 bình luận về “Giải phương trình: ` 5/(x^2 +1) + 7/(x^2+3) – (6+3x^2)/(x^2+5)=0`”

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

   $\text{→ Đặt t = x² + 1. ( t > 0 ).}$

   $\text{→ Ta có :}$

   $\text{$\dfrac{5}{t}$ + $\dfrac{7}{t + 2}$ – $\dfrac{6 + 3x²}{t + 4}$ = 0}$

   $\text{⇔ $\dfrac{5( t + 2 )( t + 4 ) + 7t( t + 4 ) – t( t + 2 )( 3x² + 6 )}{t( t + 2 )( t + 4 )}$ = 0}$

   $\text{⇒ 5( t² + 6t + 8 ) + 7t² + 28t – ( t² + 2t )( 3x² + 6 ) = 0}$

   $\text{⇔ 5t² + 30t + 40 + 7t² + 28t – ( 3x²t² + 6t² + 6x²t + 12t ) = 0}$

   $\text{⇔ 12t² + 58t + 40 – 3x²t² – 6t² – 6x²t – 12t = 0}$

   $\text{⇔ 6t² – 3x²t² – 6x²t + 46t + 40 = 0}$

   $\text{⇔ 6( x² + 1 )² – 3x²( x² + 1 )² – 6x²( x² + 1 ) + 46( x² + 1 ) + 40 = 0}$

   $\text{⇔ ( x² + 1 )²( 6 – 3x² ) – ( x² + 1 )( 6x² – 46 ) + 40 = 0}$

   $\text{⇔ ( $x^4$ + 2x² + 1 )( 6 – 3x² ) – ( $6x^4$ – 46x² + 6x² – 46 ) + 40 = 0}$

   $\text{⇔ $6x^4$ + 12x² + 6 – $3x^6$ – $6x^4$ – 3x² – $6x^4$ + 40x² + 46 + 40 = 0}$

   $\text{⇔ – $3x^6$ – $6x^4$ + 49x² + 92 = 0}$

   $\text{⇔ – $3x^6$ – $6x^5$ + $6x^5$  + $12x^4$ – $18x^4$}$

   $\text{- 36x³ + 36x³ + 72x² – 23x² – 46x + 46x + 92 = 0}$

   $\text{⇔ – $3x^5$( x + 2 ) + $6x^4$( x + 2 ) – 18x³( x + 2 )}$

   $\text{+ 36x²( x + 2 ) – 23x( x + 2 ) + 46( x + 2 ) = 0}$

   $\text{⇔ ( x + 2 )( – $3x^5$ + $6x^4$ – 18x³ + 36x² – 23x + 46 ) = 0}$

   $\text{⇔ ( x + 2 )( $3x^5$ – $6x^4$ + 18x³ – 36x² + 23x – 46 ) = 0     ( 1 )}$

   $\text{→ Xét $3x^5$ – $6x^4$ + 18x³ – 36x² + 23x – 46 = 0}$

   $\text{⇔ $3x^4$( x – 2 ) + 18x²( x – 2 ) + 23( x – 2 ) = 0}$

   $\text{⇔ ( x – 2 )( $3x^4$ + 18x² + 23 ) = 0}$

   $\text{→ Ta dễ dàng thấy : $3x^4$ + 18x² + 23 > 0  ;( $\forall$ x ).}$

   $\text{( 1 ) ⇔ ( x + 2 )( x – 2 )( $3x^4$ + 18x² + 23 ) = 0}$

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

   $\text{→ Vậy S = { 2 ; -2 }.}$

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