Giải các pt sau : `a). ({x+3}/{x-2})^2+6({x-3}/{x+2})={7(x^2-9)}/{x^2-4}` `b). (x^2+5x+6)(x^2-11x+30)=180`

1 bình luận về “Giải các pt sau : `a). ({x+3}/{x-2})^2+6({x-3}/{x+2})={7(x^2-9)}/{x^2-4}` `b). (x^2+5x+6)(x^2-11x+30)=180`”

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

   $\text{a)}$

   $\text{→ ĐKXĐ : x $\neq$ ± 2.}$

   $\text{→ Đặt :}$

   $\text{+) x + 3 = a}$

   $\text{+) x – 3 = b}$

   $\text{+) x – 2 = c}$

   $\text{+) x + 2 = d}$

   $\text{→ Ta có :}$

   $\text{$\dfrac{a²}{c²}$ + $\dfrac{6b}{d}$ = $\dfrac{7ab}{cd}$}$

   $\text{⇒ a²d + 6bc² = 7abc}$

   $\text{⇔ ( x + 3 )²( x + 2 ) + 6( x – 3 )( x – 2 )² = 7( x + 3 )( x – 3 )( x – 2 )}$

   $\text{⇔ ( x² + 6x + 9 )( x + 2 ) + 6( x – 3 )( x² – 4x + 4 ) – 7( x² – 9 )( x – 2 ) = 0}$

   $\text{⇔ x³ + 6x² + 9x + 2x² + 12x + 18 + 6( x³ – 4x² + 4x – 3x² + 12x – 12 ) – 7( x³ – 9x – 2x² + 18 ) = 0}$

   $\text{⇔ x³ + 8x² + 21x + 18 + 6( x³ – 7x² + 16x – 12 ) – 7( x³ – 2x² – 9x + 18 ) = 0}$

   $\text{⇔ x³ + 8x² + 21x + 18 + 6x³ – 42x² + 96x – 72 – 7x³ + 14x² + 63x – 126 = 0}$

   $\text{⇔ – 20x² + 180x – 180 = 0}$

   $\text{⇔ x² – 9x + 9 = 0 ⇔ x² – 9x + $\dfrac{81}{4}$ – $\dfrac{45}{4}$ = 0}$

   $\text{⇔ ( x – $\dfrac{9}{2}$ )² – $\dfrac{45}{4}$ = 0}$

   $\text{⇔ ( x – $\dfrac{9 – \sqrt{45}}{4}$ )( x – $\dfrac{9 + \sqrt{45}}{4}$ ) = 0}$

   $\text{⇔ $\left[\begin{matrix} x = \dfrac{9 – \sqrt{45}}{4}\\ x = \dfrac{9 + \sqrt{45}}{4}\end{matrix}\right.$ ( nhận ).}$

   $\text{b)}$

   $\text{→ Ta có :}$

   $\text{( x² + 5x + 6 )( x² – 11x + 30 ) = 180}$

   $\text{⇔ $x^4$ – 11x³ + 30x² + 5x³ – 55x² + 150x + 6x² – 66x + 180 = 180}$

   $\text{⇔ $x^4$ – 6x³ – 19x² + 84x = 0}$

   $\text{⇔ x( x³ – 6x² – 19x + 84 ) = 0}$

   $\text{⇔ x( x³ + 4x² – 10x² – 40x + 21x + 84 ) = 0}$

   $\text{⇔ x[ x²( x + 4 ) – 10x( x + 4 ) + 21( x + 4 ) ] = 0}$

   $\text{⇔ x( x + 4 )( x² – 10x + 21 ) = 0}$

   $\text{⇔ x( x + 4 )( x² – 7x – 3x + 21 ) = 0}$

   $\text{⇔ x( x + 4 )[ x( x – 7 ) – 3( x – 7 ) ] = 0}$

   $\text{⇔ x( x + 4 )( x – 7 )( x – 3 ) = 0}$

   $\text{⇒ S = { 0 ; – 4 ; 7 ; 3 }.}$

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