a, √2-x = √x²-2x+4 b, (x+3)√x²-4 = x² -9 Giúp mình vs ạ
1 câu trả lời
Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
2 - x \ge 0\\
{x^2} - 2x + 4 \ge 0
\end{array} \right. \Leftrightarrow x \le 2\\
\sqrt {2 - x} = \sqrt {{x^2} - 2x + 4} \\
\Leftrightarrow 2 - x = {x^2} - 2x + 4\\
\Leftrightarrow {x^2} - x + 2 = 0\\
\Leftrightarrow {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{4} = 0\\
\Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} = 0\left( {vn} \right)\\
Vậy\,x \in \emptyset \\
b)Dkxd:{x^2} - 4 \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.\\
\left( {x + 3} \right).\sqrt {{x^2} - 4} = {x^2} - 9\\
\Leftrightarrow \left( {x + 3} \right)\sqrt {{x^2} - 4} - \left( {x - 3} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right).\left( {\sqrt {{x^2} - 4} - x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\left( {tm} \right)\\
\sqrt {{x^2} - 4} = x - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
{x^2} - 4 = {x^2} - 6x + 9
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
6x = 13
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = \dfrac{{13}}{6}\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,x = - 3;x = \dfrac{{13}}{6}
\end{array}$