Giải phương trình

B

baotrant

$\left\{\begin{matrix}a=x^{2} & \\ b=\sqrt{x^{2}+2006} & \end{matrix}\right.$
(a\geq 0; b> 0)
\Rightarrow $\left\{\begin{matrix} a^{2}+b=2006 & \\ b^{2}-a=2006 & \end{matrix}\right.$
\Rightarrow $a^{2}+b=b^{2}-a$
\Leftrightarrow $(a+b)(a-b-1)=0$
\Leftrightarrow $\begin{bmatrix} a=-b & \\ a-b-1=0 & \end{bmatrix}$
\Leftrightarrow $a-1=b$
\Leftrightarrow $x^{2}-1=\sqrt{x^{2}+2006}$
\Rightarrow $x^{4}-2x^{2}+1=x^{2}+2006$
\Rightarrow $a^{2}-3a-2005=0$
 
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E

eye_smile

Cách khác:

PT \Leftrightarrow $x^4+x^2+\dfrac{1}{4}=x^2+2006-\sqrt{x^2+2006}+\dfrac{1}{4}$

\Leftrightarrow $(x^2+\dfrac{1}{2})^2=(\sqrt{x^2+2006}-\dfrac{1}{2})^2$

\Leftrightarrow $x^2+\dfrac{1}{2}=\sqrt{x^2+2006}-\dfrac{1}{2}$

\Leftrightarrow $x^2+1=\sqrt{x^2+2006}$

\Leftrightarrow $x^4+2x^2+1=x^2+2006$

Đây là PT trùng phương.
 
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