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Question

$x^{3}+ 3x^{2}-4x+1=(x^{2}+3)\sqrt{x^{2}-x+1}$

Toán Cấp 3 · Answer 1 · 5/26/2016 10:13:01 PM

$\sqrt{x^2-x+1}=\frac{x^3+3x^2-4x+1}{x^2+3}=(x+3)-\frac{7x+8}{x^2+3}$

$\Leftrightarrow \frac{7x+8}{x^2+3}+[\sqrt{x^2-x+1}-(x+3)]=0$

$\Leftrightarrow \frac{7x+8}{x^2+3}-\frac{7x+8}{\sqrt{x^2-x+1}+(x+3)}=0$

$\Leftrightarrow (7x+8)(\frac{1}{x^2+3} - \frac{1}{\sqrt{x^2-x+1}+x+3})=0$

$\Leftrightarrow x=-\frac{8}{7}$ hay

$\sqrt{x^2-x+1}=x^2-x(1)$

Đặt

$t=x^2-x$

(1) thành :

$\sqrt{t+1}=t$

$\left\{ \begin{array}{l} t\geq 0\\ t^2-t-1=0 \end{array} \right.$

$\Leftrightarrow t=\frac{1+\sqrt{5}}{2}\Rightarrow x^2-x-\frac{1+\sqrt{5}}{2}=0\Rightarrow x=\frac{1+\sqrt{3+2\sqrt{5}}}{2}$ hay

$x=\frac{1-\sqrt{3+2\sqrt{5}}}{2}$

Vậy

$x=-\frac{8}{7}$ hay

$x=\frac{1+\sqrt{3+2\sqrt{5}}}{2}$ hay

$x=\frac{1-\sqrt{3+2\sqrt{5}}}{2}$

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