Đáp án:
3) \(\left[ \begin{array}{l}
x = - 1\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:2 \ge x \ge - 2\\
x - 2 = \left( {3x - 1} \right)\sqrt {4 - {x^2}} \\
\to x - 2 - \left( {3x - 1} \right)\sqrt {\left( {2 - x} \right)\left( {x + 2} \right)} = 0\\
\to - \left( {2 - x} \right) - \left( {3x - 1} \right)\sqrt {\left( {2 - x} \right)\left( {x + 2} \right)} = 0\\
\to \left( {2 - x} \right) + \left( {3x - 1} \right)\sqrt {\left( {2 - x} \right)\left( {x + 2} \right)} = 0\\
\to \sqrt {2 - x} \left( {\sqrt {2 - x} + \left( {3x - 1} \right)\sqrt {x + 2} } \right) = 0\\
\to \left[ \begin{array}{l}
2 - x = 0\\
\sqrt {2 - x} + \left( {3x - 1} \right)\sqrt {x + 2} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
\sqrt {2 - x} = - \left( {3x - 1} \right)\sqrt {x + 2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
\sqrt {2 - x} = \left( {1 - 3x} \right)\sqrt {x + 2} \left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to 2 - x = \left( {1 - 6x + 9{x^2}} \right)\left( {x + 2} \right)\\
\to 9{x^3} + 18{x^2} - 6{x^2} - 12x + x + 2 - 2 + x = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 2 + \sqrt {14} }}{3}\left( {TM} \right)\\
x = \dfrac{{ - 2 - \sqrt {14} }}{3}\left( {TM} \right)\\
x = 0\left( {TM} \right)
\end{array} \right.\\
KL:\left[ \begin{array}{l}
x = \dfrac{{ - 2 + \sqrt {14} }}{3}\\
x = \dfrac{{ - 2 - \sqrt {14} }}{3}\\
x = 0\\
x = 2
\end{array} \right.\\
2)DK:x \ge 0\\
2 - \sqrt {2 + x} = {x^2}\\
\to 2 - {x^2} = \sqrt {2 + x} \\
\to 4 - 4{x^2} + {x^4} = 2 + x\\
\to {x^4} - 4{x^2} - x + 2 = 0\\
\to {x^4} - 2{x^3} + 2{x^3} - 4{x^2} - x + 2 = 0\\
\to {x^3}\left( {x - 2} \right) + 2{x^2}\left( {x - 2} \right) - \left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {{x^3} + 2{x^2} - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x - 2 = 0\\
{x^3} + {x^2} + {x^2} + x - x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
\left( {x + 1} \right)\left( {{x^2} + x - 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt 5 }}{2}\\
x = \dfrac{{ - 1 - \sqrt 5 }}{2}\left( l \right)\\
x = - 1\left( l \right)\\
x = 2
\end{array} \right.\\
3)DK:x \ge - 1\\
\sqrt {x + 1 + 2\sqrt {x + 1} .1 + 1} + \sqrt {x + 1 - 2\sqrt {x + 1} .1 + 1} = \dfrac{{x + 5}}{2}\\
\to \sqrt {{{\left( {\sqrt {x + 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x + 1} - 1} \right)}^2}} = \dfrac{{x + 5}}{2}\\
\to \sqrt {x + 1} + 1 + \left| {\sqrt {x + 1} - 1} \right| = \dfrac{{x + 5}}{2}\\
\to \left[ \begin{array}{l}
\sqrt {x + 1} + 1 + \sqrt {x + 1} - 1 = \dfrac{{x + 5}}{2}\left( {DK:x \ge 2} \right)\\
\sqrt {x + 1} + 1 - \sqrt {x + 1} + 1 = \dfrac{{x + 5}}{2}\left( {DK: - 1 \le x < 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\sqrt {x + 1} = \dfrac{{x + 5}}{2}\\
2 = \dfrac{{x + 5}}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt {x + 1} = \dfrac{{x + 5}}{4}\\
x + 5 = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x + 1 = \dfrac{{{x^2} + 10x + 25}}{{16}}\\
x = - 1\left( {TM} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
{x^2} + 10x + 25 = 16x + 16
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 3\left( {TM} \right)
\end{array} \right.
\end{array}\)
