X^4-12x^2+16=0 Mong các bn giúp mik giải bài này ạ 09/09/2021 Bởi Alice X^4-12x^2+16=0 Mong các bn giúp mik giải bài này ạ
$x^4-12x^2+16=0$ $⇔(x^4-12x^2+6^2) -20=0$ $⇔(x^2-6)^2 = 20$ $⇔\left[ \begin{array}{l}x^2-6=\sqrt[]{20}\\x^2-6=-\sqrt[]{20}\end{array} \right.$ $⇔\left[ \begin{array}{l}x^2=\sqrt[]{20}+6\\x^2=6-\sqrt[]{20}\end{array} \right.$ $⇔\left[ \begin{array}{l}x=±\sqrt[]{\sqrt[]{20}+6}\\x=±\sqrt[]{6-\sqrt[]{20}}\end{array} \right.$ $⇔\left[ \begin{array}{l}x=±(\sqrt[]{5}+1)\\x=±(\sqrt[]{5}-1)\end{array} \right.$ Bình luận
$x^4-12x^2+16=0$
$⇔(x^4-12x^2+6^2) -20=0$
$⇔(x^2-6)^2 = 20$
$⇔\left[ \begin{array}{l}x^2-6=\sqrt[]{20}\\x^2-6=-\sqrt[]{20}\end{array} \right.$
$⇔\left[ \begin{array}{l}x^2=\sqrt[]{20}+6\\x^2=6-\sqrt[]{20}\end{array} \right.$
$⇔\left[ \begin{array}{l}x=±\sqrt[]{\sqrt[]{20}+6}\\x=±\sqrt[]{6-\sqrt[]{20}}\end{array} \right.$
$⇔\left[ \begin{array}{l}x=±(\sqrt[]{5}+1)\\x=±(\sqrt[]{5}-1)\end{array} \right.$