\Leftrightarrow[TEX]\sqrt{x^2 + 3x +3} = 3 - \sqrt{x^2 - 3x +6}[/TEX]
\Leftrightarrow[TEX] (\sqrt{x^2 + 3x +3} )^2= (3 - \sqrt{x^2 - 3x +6})^2[/TEX]
\Leftrightarrow [TEX] x^2 + 3x + 3 = 9 - 6 \sqrt{x^2 - 3x +6} + x^2 - 3x +6[/TEX]
\Leftrightarrow[TEX]x^2 + 3x +3 - 9 -x^2 +3x - 6 = - 6 \sqrt{x^2 - 3x +6}[/TEX]
\Leftrightarrow[TEX]6x - 3 =- 6 \sqrt{x^2 - 3x +6} [/TEX]
\Leftrightarrow[TEX]6 \sqrt{x^2 - 3x +6}= 3- 6x[/TEX]đièu kiẹn
\Leftrightarrow[TEX]\left\{ \begin{array}{l} 12-6x >=0 \\ (6 \sqrt{x^2 - 3x +6})^2= (12- 6x)^2 \end{array} \right[/TEX]
\Leftrightarrow[TEX]\left\{ \begin{array}{l} x>=2\\ (6 \sqrt{x^2 - 3x +6})^2= (12- 6x)^2 \end{array} \right[/TEX]
[TEX]\left\{ \begin{array}{l} x>=2 \\ 36x=72 \end{array} \right[/TEX]
[TEX]\left\{ \begin{array}{l} x>=2 \\ x=2(TM) \end{array} \right[/TEX]
vậy x=2
uh, bạn tính nhầm. wên mất ở đó còn có 9 wen chưa trừ!!!