Toán Giúp mk giải 1 bài tìm nghiệm 2x^4+16x^3-2x^2-124x+126=0 18/09/2021 By Ivy Giúp mk giải 1 bài tìm nghiệm 2x^4+16x^3-2x^2-124x+126=0
$2x^4+16x^3-2x^2-124x+126=0$ $2(x^4+8x^3-x^2-62x+63)=0$ $⇔x^4+5x^3-9x^2+3x^3+15x^2-27x-7x^2-35x+63=0$ $⇔x^2(x^2+5x-9)+3x(x^2+5x-9)-7(x^2-5x+9)=0$ $⇔(x^2+5x-9)(x^2+3x-7)=0$ $⇔$\(\left[ \begin{array}{l}x^2+5x-9=0\\x^2+3x-7=0\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}=\dfrac{71}{4}\\x^2+2.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{37}{4}\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}(x+\dfrac{5}{2})^2=\dfrac{71}{4}\\(x+\dfrac{3}{2})^2=\dfrac{37}{4}\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x+\dfrac{5}{2}=\dfrac{\sqrt[]{71}}{2}\\x+\dfrac{5}{2}=\dfrac{-\sqrt[]{71}}{2}\\x+\dfrac{3}{2}=\dfrac{\sqrt[]{37}}{2}\\x+\dfrac{3}{2}=-\dfrac{\sqrt[]{37}}{2}\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x=\dfrac{\sqrt[]{71}-5}{2}\\x=\dfrac{-\sqrt[]{71}-5}{2}\\x=\dfrac{\sqrt[]{37}-3}{2}\\x=-\dfrac{\sqrt[]{37}-3}{2}\end{array} \right.\) Trả lời
$2x^4+16x^3-2x^2-124x+126=0$
$2(x^4+8x^3-x^2-62x+63)=0$
$⇔x^4+5x^3-9x^2+3x^3+15x^2-27x-7x^2-35x+63=0$
$⇔x^2(x^2+5x-9)+3x(x^2+5x-9)-7(x^2-5x+9)=0$
$⇔(x^2+5x-9)(x^2+3x-7)=0$
$⇔$\(\left[ \begin{array}{l}x^2+5x-9=0\\x^2+3x-7=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}=\dfrac{71}{4}\\x^2+2.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{37}{4}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}(x+\dfrac{5}{2})^2=\dfrac{71}{4}\\(x+\dfrac{3}{2})^2=\dfrac{37}{4}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x+\dfrac{5}{2}=\dfrac{\sqrt[]{71}}{2}\\x+\dfrac{5}{2}=\dfrac{-\sqrt[]{71}}{2}\\x+\dfrac{3}{2}=\dfrac{\sqrt[]{37}}{2}\\x+\dfrac{3}{2}=-\dfrac{\sqrt[]{37}}{2}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{\sqrt[]{71}-5}{2}\\x=\dfrac{-\sqrt[]{71}-5}{2}\\x=\dfrac{\sqrt[]{37}-3}{2}\\x=-\dfrac{\sqrt[]{37}-3}{2}\end{array} \right.\)