(x-1)(x-3)(x+5)(x+7)-297=0 giải giúp mik với

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1. $(x-1)(x-3)(x+5)(x+7)-297=0$

   $⇔[(x-1)(x+5)].[(x-3)(x+7)]=0$

   $⇔(x^2+4x-5)(x^2+4x-21)=0$

   Đặt: $y=x^2+4x-13$

   $⇒(y+8)(y-8)-297=0$

   $⇔y^2-64-297=0$

   $⇔y^2-361=0$

   $⇔(y-19)(y+19)=0$

   $⇔(x^2+4x-32)(x^2+4x-2)=0$

   $⇔(x-4)(x+8)(x+2-$$\sqrt[]{6})(x+2+$ $\sqrt[]{6})=0$

   $⇔$ \(\left[ \begin{array}{l}x-4=0\\x+8=0\\x+2-\sqrt[]{6}=0\\x+2+ \sqrt[]{6}=0\end{array} \right.\)

   $⇔$\(\left[ \begin{array}{l}x=4\\x=-8\\x=-2+\sqrt[]{6}\\x=-2-\sqrt[]{6}\end{array} \right.\)  

   Vậy $S=\{4;-8;-2+\sqrt[]{6};-2-\sqrt[]{6}\}$

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