Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
5,\\
a,\\
\sqrt {14 + 6\sqrt 5 } + \sqrt {14 - 6\sqrt 5 } \\
= \sqrt {9 + 2.3.\sqrt 5 + 5} + \sqrt {9 - 2.3.\sqrt 5 + 5} \\
= \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} + \sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} \\
= 3 + \sqrt 5 + 3 - \sqrt 5 \\
= 6\\
b,\\
\sqrt {6 + 4\sqrt 2 } + \sqrt {11 - 6\sqrt 2 } \\
= \sqrt {4 + 2.2.\sqrt 2 + 2} + \sqrt {9 - 2.3.\sqrt 2 + 2} \\
= \sqrt {{{\left( {2 + \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} \\
= 2 + \sqrt 2 + 3 - \sqrt 2 \\
= 5\\
c,\\
\sqrt {29 + 12\sqrt 5 } - \sqrt {29 - 12\sqrt 5 } \\
= \sqrt {20 + 2.2\sqrt 5 .3 + 9} - \sqrt {20 - 2.2\sqrt 5 .3 + 9} \\
= \sqrt {{{\left( {2\sqrt 5 + 3} \right)}^2}} - \sqrt {{{\left( {2\sqrt 5 - 3} \right)}^2}} \\
= \left( {2\sqrt 5 + 3} \right) - \left( {2\sqrt 5 - 3} \right)\\
= 6\\
d,\\
\sqrt {32 + 10\sqrt 7 } + \sqrt {32 - 10\sqrt 7 } \\
= \sqrt {25 + 2.5.\sqrt 7 + 7} + \sqrt {25 - 2.5.\sqrt 7 + 7} \\
= \sqrt {{{\left( {5 + \sqrt 7 } \right)}^2}} + \sqrt {{{\left( {5 - \sqrt 7 } \right)}^2}} \\
= 5 + \sqrt 7 + 5 - \sqrt 7 \\
= 10\\
6,\\
a,\\
5 + \sqrt 5 = \sqrt 5 .\left( {\sqrt 5 + 1} \right)\\
b,\\
a - 2\sqrt a = \sqrt a \left( {\sqrt a - 2} \right)\\
c,\\
x - \sqrt {xy} = \sqrt x \left( {\sqrt x - \sqrt y } \right)\\
d,\\
x\sqrt y - y\sqrt x = {\sqrt x ^2}.\sqrt y - {\sqrt y ^2}.\sqrt x = \sqrt {xy} .\left( {\sqrt x - \sqrt y } \right)\\
e,\\
x - y - \sqrt x - \sqrt y = \left( {x - y} \right) - \left( {\sqrt x + \sqrt y } \right)\\
= \left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right) - \left( {\sqrt x + \sqrt y } \right)\\
= \left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x - \sqrt y - 1} \right)\\
g,\\
1 - x = {1^2} - {\sqrt x ^2} = \left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)\\
h,\\
1 - 2\sqrt x + x = {\sqrt x ^2} - 2.\sqrt x .1 + {1^2} = {\left( {\sqrt x - 1} \right)^2}\\
k,\\
x - 5\sqrt x + 6 = \left( {x - 3\sqrt x } \right) - \left( {2\sqrt x - 6} \right)\\
= \sqrt x \left( {\sqrt x - 3} \right) - 2.\left( {\sqrt x - 3} \right)\\
= \left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)\\
i,\\
x - 2\sqrt x - 3 = \left( {x - 3\sqrt x } \right) + \left( {\sqrt x - 3} \right)\\
= \sqrt x \left( {\sqrt x - 3} \right) + \left( {\sqrt x - 3} \right)\\
= \left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)
\end{array}\)
