Đáp án:
\(\begin{array}{l}
10)\sqrt {x + 1} - \sqrt {x - 2} \\
12)\sqrt {x + 3} - \sqrt {x - 1}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
10)DK:\left[ \begin{array}{l}
x \ge 2\\
x \le - 1
\end{array} \right.\\
A = \sqrt {2x - 1 - 2\sqrt {{x^2} - x - 2} } \\
= \sqrt {2x - 1 - 2\sqrt {\left( {x - 2} \right)\left( {x + 1} \right)} } \\
= \sqrt {x + 1 - 2\sqrt {\left( {x - 2} \right)\left( {x + 1} \right)} + x - 2} \\
= \sqrt {{{\left( {\sqrt {x + 1} - \sqrt {x - 2} } \right)}^2}} \\
= \sqrt {x + 1} - \sqrt {x - 2} \\
12)DK:\left[ \begin{array}{l}
x \ge 1\\
x \le - 3
\end{array} \right.\\
\sqrt {2x + 2 - 2\sqrt {{x^2} + 2x - 3} } \\
= \sqrt {2x + 2 - 2\sqrt {\left( {x + 3} \right)\left( {x - 1} \right)} } \\
= \sqrt {x + 3 - 2\sqrt {\left( {x + 3} \right)\left( {x - 1} \right)} + x - 1} \\
= \sqrt {{{\left( {\sqrt {x + 3} - \sqrt {x - 1} } \right)}^2}} \\
= \sqrt {x + 3} - \sqrt {x - 1}
\end{array}\)
