GPT: $x +$ $\sqrt[]{4-x^2}$ $= 2 + 3x$ $\sqrt[]{4-x^2}$

Đáp án:

$\begin{array}{l}
Dkxd:4 – {x^2} \ge 0\\
 \Rightarrow  – 2 \le x \le 2\\
Đặt:x + \sqrt {4 – {x^2}}  = t\\
 \Rightarrow {x^2} + 2.x.\sqrt {4 – {x^2}}  + 4 – {x^2} = {t^2}\\
 \Rightarrow x.\sqrt {4 – {x^2}}  = \dfrac{{{t^2} – 4}}{2}\\
 \Rightarrow 3.x.\sqrt {4 – {x^2}}  = \dfrac{{3{t^2} – 12}}{2}\\
Pt: \Rightarrow t = 2 + \dfrac{{3{t^2} – 12}}{2}\\
 \Rightarrow 3{t^2} – 2t – 8 = 0\\
 \Rightarrow 3{t^2} – 6t + 4t – 8 = 0\\
 \Rightarrow \left( {t – 2} \right)\left( {3t + 4} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
t = 2\\
t =  – \dfrac{4}{3}
\end{array} \right.\\
 + Khi:t = 2\\
 \Rightarrow x.\sqrt {4 – {x^2}}  = \dfrac{{{t^2} – 4}}{2} = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x =  – 2
\end{array} \right.\left( {tmdk} \right)\\
 + Khi:t =  – \dfrac{4}{3}\\
 \Rightarrow x.\sqrt {4 – {x^2}}  = \dfrac{{{t^2} – 4}}{2} = \dfrac{{ – 10}}{9}\\
 \Rightarrow \left\{ \begin{array}{l}
 – 2 \le x < 0\\
{x^2}\left( {4 – {x^2}} \right) = \dfrac{{100}}{{81}}
\end{array} \right.\\
 \Rightarrow {x^4} – 4{x^2} + \dfrac{{100}}{{81}} = 0\\
 \Rightarrow {x^2} = \dfrac{{22 + \sqrt {14} }}{9}\left( {ktm} \right)\\
Vậy\,x \in \left\{ {0; – 2;2} \right\}
\end{array}$