Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
a,\\
A = \sqrt {x - 1 + 2\sqrt {x - 2} } + \sqrt {x - 1 - 2\sqrt {x - 2} } \\
= \sqrt {\left( {x - 2} \right) + 2\sqrt {x - 2} + 1} + \sqrt {\left( {x - 2} \right) - 2\sqrt {x - 2} + 1} \\
= \sqrt {{{\sqrt {x - 2} }^2} + 2\sqrt {x - 2} .1 + {1^2}} + \sqrt {{{\sqrt {x - 2} }^2} - 2.\sqrt {x - 2} .1 + {1^2}} \\
= \sqrt {{{\left( {\sqrt {x - 2} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 2} - 1} \right)}^2}} \\
= \left| {\sqrt {x - 2} + 1} \right| + \left| {\sqrt {x - 2} - 1} \right|\\
x \ge 3 \Rightarrow \sqrt {x - 2} \ge 1 \Rightarrow \sqrt {x - 2} - 1 \ge 0\\
\sqrt {x - 2} + 1 \ge 1 > 0\\
\Rightarrow A = \left( {\sqrt {x - 2} + 1} \right) + \left( {\sqrt {x - 2} - 1} \right) = 2\sqrt {x - 2} \\
b,\\
A = \sqrt {x + 3 + 4\sqrt {x - 1} } + \sqrt {x + 3 - 4\sqrt {x - 1} } \\
= \sqrt {\left( {x - 1} \right) + 4\sqrt {x - 1} + 4} + \sqrt {\left( {x - 1} \right) - 4\sqrt {x - 1} + 4} \\
= \sqrt {{{\sqrt {x - 1} }^2} + 2.\sqrt {x - 1} .2 + {2^2}} + \sqrt {{{\sqrt {x - 1} }^2} - 2.\sqrt {x - 1} .2 + {2^2}} \\
= \sqrt {{{\left( {\sqrt {x - 1} + 2} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 2} \right)}^2}} \\
= \left| {\sqrt {x - 1} + 2} \right| + \left| {\sqrt {x - 1} - 2} \right|\\
1 \le x < 5 \Rightarrow 0 \le x - 1 < 4 \Leftrightarrow 0 \le \sqrt {x - 1} < 2 \Rightarrow \sqrt {x - 1} - 2 < 0\\
\Rightarrow \left| {\sqrt {x - 1} - 2} \right| = 2 - \sqrt {x - 1} \\
\sqrt {x - 1} + 2 \ge 2 > 0\\
\Rightarrow A = \left( {\sqrt {x - 1} + 2} \right) + \left( {2 - \sqrt {x - 1} } \right) = 4\\
4,\\
a,\\
DKXD:\,\,\,{x^2} - 2x + 4 \ge 0,\,\,\,\forall x\\
\sqrt {{x^2} - 2x + 4} = 2x - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 2 \ge 0\\
{x^2} - 2x + 4 = {\left( {2x - 2} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
{x^2} - 2x + 4 = 4{x^2} - 8x + 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
3{x^2} - 6x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
3x\left( {x - 2} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 2\\
b,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x - 3 \ge 0\\
{x^2} - 9 \ge 0
\end{array} \right. \Leftrightarrow x \ge 3\\
\sqrt {x - 3} - 2\sqrt {{x^2} - 9} = 0\\
\Leftrightarrow \sqrt {x - 3} - 2\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} = 0\\
\Leftrightarrow \sqrt {x - 3} \left( {1 - 2\sqrt {x + 3} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 3} = 0\\
1 - 2\sqrt {x + 3} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
\sqrt {x + 3} = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{{11}}{4}\,\,\,\,\left( {L,\,\,\,x \ge 3} \right)
\end{array} \right.\\
\Rightarrow x = 3\\
c,\\
DKXD:\,\,\,x \ge 1\\
\sqrt {x + 3 - 4\sqrt {x - 1} } = 3\\
\Leftrightarrow \sqrt {\left( {x - 1} \right) - 4\sqrt {x - 1} + 4} = 3\\
\Leftrightarrow \sqrt {{{\sqrt {x - 1} }^2} - 2.\sqrt {x - 1} .2 + {2^2}} = 3\\
\Leftrightarrow \sqrt {{{\left( {\sqrt {x - 1} - 2} \right)}^2}} = 3\\
\Leftrightarrow \left| {\sqrt {x - 1} - 2} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 1} - 2 = 3\\
\sqrt {x - 1} - 2 = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 1} = 5\\
\sqrt {x - 1} = - 1\,\,\,\,\left( L \right)
\end{array} \right.\\
\Leftrightarrow x - 1 = 25\\
\Leftrightarrow x = 26
\end{array}\)
