Đáp án:
$\begin{array}{l}
1D = \sqrt {3 - \sqrt 5 } .\left( {\sqrt {10} + \sqrt 2 } \right)\\
= \sqrt {3 - \sqrt 5 } .\sqrt 2 \left( {\sqrt 5 + 1} \right)\\
= \sqrt {6 - 2\sqrt 5 } .\left( {\sqrt 5 + 1} \right)\\
= \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} .\left( {\sqrt 5 + 1} \right)\\
= \left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)\\
= 5 - 1\\
= 4\\
2)\\
c)Dkxd:x \ge 3\\
\sqrt {\dfrac{{x - 3}}{4}} + 2\sqrt {9x - 27} = \sqrt {x - 3} + 1\\
\Leftrightarrow \dfrac{1}{2}\sqrt {x - 3} + 2.3\sqrt {x - 3} = \sqrt {x - 3} + 1\\
\Leftrightarrow \dfrac{{11}}{2}\sqrt {x - 3} = 1\\
\Leftrightarrow \sqrt {x - 3} = \dfrac{2}{{11}}\\
\Leftrightarrow x - 3 = \dfrac{4}{{121}}\\
\Leftrightarrow x = \dfrac{{367}}{{121}}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{367}}{{121}}\\
d)Dkxd:1 - 20x \ge 0\\
\Leftrightarrow x \le \dfrac{1}{{20}}\\
\sqrt {{x^2} - 2x + 1} = 1 - 20x\\
\Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}} = 1 - 20x\\
\Leftrightarrow 1 - x = 1 - 20x\left( {do:x \le \dfrac{1}{{20}}} \right)\\
\Leftrightarrow x = 0\left( {tm} \right)\\
Vậy\,x = 0\\
e)Dkxd:x \ge 1\\
\sqrt {x + 3 - 4\sqrt {x - 1} } + \sqrt {x + 8 + 6\sqrt {x - 1} } = 1\\
\Leftrightarrow \sqrt {x - 1 - 4\sqrt {x - 1} + 4} + \sqrt {x - 1 + 6\sqrt {x - 1} + 9} = 1\\
\Leftrightarrow \sqrt {{{\left( {\sqrt {x - 1} - 2} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} + 3} \right)}^2}} = 1\\
\Leftrightarrow \left| {\sqrt {x - 1} - 2} \right| + \sqrt {x - 1} + 3 = 1\\
+ Khi:\sqrt {x - 1} - 2 \ge 0 \Leftrightarrow x \ge 5\\
\Leftrightarrow \sqrt {x - 1} - 2 + \sqrt {x - 1} + 3 = 1\\
\Leftrightarrow \sqrt {x - 1} = 0\left( {ktm} \right)\\
+ Khi:\sqrt {x - 1} - 2 < 0 \Leftrightarrow 1 \le x < 5\\
\Leftrightarrow 2 - \sqrt {x - 1} + \sqrt {x - 1} + 3 = 1\\
\Leftrightarrow 5 = 1\left( {ktm} \right)
\end{array}$
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