Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
A = \sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } \,\,\,\,\,\,\,\,\left( {x \ge 1} \right)\\
= \sqrt {\left( {x - 1} \right) + 2.\sqrt {x - 1} + 1} + \sqrt {\left( {x - 1} \right) - 2\sqrt {x - 1} + 1} \\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} \\
= \left| {\sqrt {x - 1} + 1} \right| + \left| {\sqrt {x - 1} - 1} \right|\\
*)\,\,\,TH1:\,\,\,\,\sqrt {x - 1} > 1 \Leftrightarrow x > 2 \Rightarrow \left| {\sqrt {x - 1} - 1} \right| = \sqrt {x - 1} - 1\\
\Rightarrow A = \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1 = 2\sqrt {x - 1} \\
*)\,\,\,\,\,TH2:\,\,\,\sqrt {x - 1} \le 1 \Leftrightarrow 1 \le x \le 2 \Rightarrow \left| {\sqrt {x - 1} - 1} \right| = 1 - \sqrt {x - 1} \\
\Rightarrow A = \sqrt {x - 1} + 1 + 1 - \sqrt {x - 1} = 2\\
2,\\
A = \left( {b - 2} \right).\sqrt {\dfrac{{{b^2}}}{{4 - 4b + {b^2}}}} \\
= \left( {b - 2} \right).\sqrt {\dfrac{{{b^2}}}{{{{\left( {2 - b} \right)}^2}}}} \\
0 < b < 2 \Rightarrow 2 - b > 0 \Rightarrow \dfrac{b}{{2 - b}} > 0\\
\Rightarrow A = \left( {b - 2} \right).\sqrt {{{\left( {\dfrac{b}{{2 - b}}} \right)}^2}} = \left( {b - 2} \right).\dfrac{b}{{2 - b}} = - b\\
B = \dfrac{{x - y}}{{\sqrt x - \sqrt y }}:\left( {\sqrt x - \sqrt y } \right)\,\,\,\,\,\,\,\,\,\,\left( {x,y \ge 0,\,\,\,x \ne y} \right)\\
= \dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x - \sqrt y }}.\dfrac{1}{{\sqrt x - \sqrt y }}\\
= \dfrac{{\sqrt x + \sqrt y }}{{\sqrt x - \sqrt y }}\\
3,\\
x + y + 12 = 4\sqrt x + 6\sqrt {y - 1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0;\,\,y \ge 1} \right)\\
\Leftrightarrow x + y + 12 - 4\sqrt x - 6\sqrt {y - 1} = 0\\
\Leftrightarrow \left( {x - 4\sqrt x + 4} \right) + \left[ {\left( {y - 1} \right) - 6\sqrt {y - 1} + 9} \right] = 0\\
\Leftrightarrow {\left( {\sqrt x - 2} \right)^2} + {\left( {\sqrt {y - 1} - 3} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x - 2 = 0\\
\sqrt {y - 1} - 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt x = 2\\
\sqrt {y - 1} = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = 10
\end{array} \right.
\end{array}\)
