(x-1)(x-3)(x+5)(x+7)-297=0 giải giúp mik với 29/11/2021 Bởi Melody (x-1)(x-3)(x+5)(x+7)-297=0 giải giúp mik với
$(x-1)(x-3)(x+5)(x+7)-297=0$ $⇔[(x-1)(x+5)].[(x-3)(x+7)]=0$ $⇔(x^2+4x-5)(x^2+4x-21)=0$ 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 $S=\{4;-8;-2+\sqrt[]{6};-2-\sqrt[]{6}\}$ Bình luận
$(x-1)(x-3)(x+5)(x+7)-297=0$
$⇔[(x-1)(x+5)].[(x-3)(x+7)]=0$
$⇔(x^2+4x-5)(x^2+4x-21)=0$
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 $S=\{4;-8;-2+\sqrt[]{6};-2-\sqrt[]{6}\}$