Đáp án:
\(A8 = \sqrt {2x - 2} \)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge \frac{1}{2}\\
A8 = \left\{ {\frac{{\sqrt {\left[ {{{\left( {\sqrt {x - 1} } \right)}^2} + 2\sqrt {x - 1} + 1} \right]} + \sqrt {\left[ {{{\left( {\sqrt {x - 1} } \right)}^2} - 2\sqrt {x - 1} + 1} \right]} }}{{\sqrt {\left[ {{{\left( {\sqrt {x - \frac{1}{2}} } \right)}^2} + 2.\frac{1}{2}.\sqrt 2 .\sqrt {x - \frac{1}{2}} + {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}} \right]} + \sqrt {\left[ {{{\left( {\sqrt {x - \frac{1}{2}} } \right)}^2} - 2.\frac{1}{2}.\sqrt 2 .\sqrt {x - \frac{1}{2}} + {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}} \right]} }}} \right\}.\sqrt {2x - 1} \\
= \frac{{\sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} }}{{\sqrt {{{\left( {\sqrt {x - \frac{1}{2}} + \frac{{\sqrt 2 }}{2}} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - \frac{1}{2}} - \frac{{\sqrt 2 }}{2}} \right)}^2}} }}.\sqrt {2x - 1} \\
= \frac{{\sqrt {x - 1} + 1 + \sqrt {x - 1} - 1}}{{\sqrt {x - \frac{1}{2}} + \frac{{\sqrt 2 }}{2} + \sqrt {x - \frac{1}{2}} - \frac{{\sqrt 2 }}{2}}}.\sqrt {2x - 1} \\
= \frac{{2\sqrt {x - 1} .\sqrt {2x - 1} }}{{2\sqrt {x - \frac{1}{2}} }} = \frac{{\sqrt 2 .\sqrt {x - 1} .\sqrt {2x - 1} }}{{\sqrt {2x - 1} }}\\
= \sqrt {2x - 2}
\end{array}\)
