Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
2,\\
a,\\
\sqrt {5 + 2\sqrt 6 }  - \sqrt {5 - 2\sqrt 6 }  = \sqrt {3 + 2.\sqrt 3 .\sqrt 2  + 2}  - \sqrt {3 - 2.\sqrt 3 .\sqrt 2  + 2} \\
 = \sqrt {{{\left( {\sqrt 3  + \sqrt 2 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 3  - \sqrt 2 } \right)}^2}}  = \left| {\sqrt 3  + \sqrt 2 } \right| - \left| {\sqrt 3  - \sqrt 2 } \right|\\
 = \left( {\sqrt 3  + \sqrt 2 } \right) - \left( {\sqrt 3  - \sqrt 2 } \right) = 2\sqrt 2 \\
b,\\
\sqrt {8 + \sqrt {60} }  - \sqrt {8 - 2\sqrt {15} }  = \sqrt {8 + 2\sqrt {15} }  - \sqrt {8 - 2\sqrt {15} } \\
 = \sqrt {5 + 2.\sqrt 5 .\sqrt 3  + 3}  - \sqrt {5 - 2.\sqrt 5 .\sqrt 3  + 3} \\
 = \sqrt {{{\left( {\sqrt 5  + \sqrt 3 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 5  - \sqrt 3 } \right)}^2}}  = \left| {\sqrt 5  + \sqrt 3 } \right| - \left| {\sqrt 5  - \sqrt 3 } \right|\\
 = \left( {\sqrt 5  + \sqrt 3 } \right) - \left( {\sqrt 5  - \sqrt 3 } \right) = 2\sqrt 3 \\
c,\\
\sqrt {4 - \sqrt 7 }  + \sqrt {4 + \sqrt 7 }  = \sqrt {\dfrac{1}{2}.\left( {8 - 2\sqrt 7 } \right)}  - \sqrt {\dfrac{1}{2}.\left( {8 + 2\sqrt 7 } \right)} \\
 = \sqrt {\dfrac{1}{2}.\left( {7 - 2.\sqrt 7 .1 + 1} \right)}  - \sqrt {\dfrac{1}{2}.\left( {7 + 2.\sqrt 7 .1 + 1} \right)} \\
 = \sqrt {\dfrac{1}{2}.{{\left( {\sqrt 7  - 1} \right)}^2}}  - \sqrt {\dfrac{1}{2}.{{\left( {\sqrt 7  + 1} \right)}^2}} \\
 = \dfrac{1}{{\sqrt 2 }}.\left| {\sqrt 7  - 1} \right| - \dfrac{1}{{\sqrt 2 }}.\left| {\sqrt 7  + 1} \right|\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 7  - 1} \right) - \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 7  + 1} \right)\\
 =  - \dfrac{2}{{\sqrt 2 }} =  - \sqrt 2 \\
d,\\
\sqrt {5\sqrt 3  + 5\sqrt {48 - 10\sqrt {7 + 4\sqrt 3 } } } \\
 = \sqrt {5\sqrt 3  + 5\sqrt {48 - 10\sqrt {4 + 2.2.\sqrt 3  + 3} } } \\
 = \sqrt {5\sqrt 3  + 5\sqrt {48 - 10\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} } } \\
 = \sqrt {5\sqrt 3  + 5\sqrt {48 - 10\left( {2 + \sqrt 3 } \right)} } \\
 = \sqrt {5\sqrt 3  + 5\sqrt {28 - 10\sqrt 3 } } \\
 = \sqrt {5\sqrt 3  + 5.\sqrt {25 - 2.5.\sqrt 3  + 3} } \\
 = \sqrt {5\sqrt 3  + 5.\sqrt {{{\left( {5 - \sqrt 3 } \right)}^2}} } \\
 = \sqrt {5\sqrt 3  + 5.\left( {5 - \sqrt 3 } \right)} \\
 = \sqrt {25}  = 5\\
e,\\
\left( {x - 4} \right).\sqrt {16 - 8x + {x^2}} \\
 = \left( {x - 4} \right).\sqrt {{4^2} - 2.4.x + {x^2}} \\
 = \left( {x - 4} \right).\sqrt {{{\left( {4 - x} \right)}^2}} \\
 = \left( {x - 4} \right).\left| {4 - x} \right|\\
x \ge 4 \Rightarrow \left| {4 - x} \right| = x - 4\\
 \Rightarrow \left( {x - 4} \right).\sqrt {16 - 8x + {x^2}}  = \left( {x - 4} \right).\left( {x - 4} \right) = {\left( {x - 4} \right)^2}\\
f,\\
F = \left( {2x - 5} \right).\sqrt {\dfrac{2}{{{{\left( {2x - 5} \right)}^2}}}} \\
 = \left( {2x - 5} \right).\dfrac{{\sqrt 2 }}{{\left| {2x - 5} \right|}}\\
TH1:\,\,\,2x - 5 > 0 \Leftrightarrow x > \dfrac{5}{2}\\
 \Rightarrow \left| {2x - 5} \right| = 2x - 5\\
 \Rightarrow F = \sqrt 2 \\
TH2:\,\,\,\,2x - 5 < 0 \Leftrightarrow x < \dfrac{5}{2}\\
 \Leftrightarrow \left| {2x - 5} \right| =  - \left( {2x - 5} \right)\\
 \Rightarrow F =  - \sqrt 2 \\
g,\\
\sqrt {x - 4\sqrt {x - 4} } \\
 = \sqrt {\left( {x - 4} \right) - 4.\sqrt {x - 4}  + 4} \\
 = \sqrt {{{\sqrt {x - 4} }^2} - 2.\sqrt {x - 4} .2 + {2^2}} \\
 = \sqrt {{{\left( {\sqrt {x - 4}  - 2} \right)}^2}} \\
 = \left| {\sqrt {x - 4}  - 2} \right|
\end{array}\)

\(\begin{array}{l}
5,\\
a)\\
\sqrt {0,16{x^2}}  = \sqrt {0,{4^2}.{x^2}}  = \sqrt {{{\left( {0,4x} \right)}^2}}  = \left| {0,4x} \right| = 0,4.\left| x \right| =  - 0,4x\\
\left( {x < 0 \Rightarrow \left| x \right| =  - x} \right)\\
b,\\
\sqrt {12.75.{{\left( {2 - x} \right)}^2}}  = \sqrt {{2^2}{{.3.3.5}^2}.{{\left( {2 - x} \right)}^2}}  = \sqrt {{{\left( {2.3.5} \right)}^2}.{{\left( {2 - x} \right)}^2}} \\
 = 30.\left| {2 - x} \right| = 30.\left( {2 - x} \right) = 60 - 30x\\
\left( {x < 2 \Rightarrow 2 - x > 0 \Rightarrow \left| {2 - x} \right| = 2 - x} \right)\\
c,\\
\sqrt {\dfrac{x}{{15}}} .\sqrt {\dfrac{{5x}}{3}}  = \sqrt {\dfrac{x}{{15}}.\dfrac{{5x}}{3}}  = \sqrt {\dfrac{{{x^2}}}{9}}  = \sqrt {{{\left( {\dfrac{x}{3}} \right)}^2}}  = \left| {\dfrac{x}{3}} \right| = \dfrac{x}{3}\\
\left( {x > 0 \Rightarrow \left| x \right| = x} \right)\\
d,\\
\sqrt {3x} .\sqrt {\dfrac{{12}}{x}}  = \sqrt {3x.\dfrac{{12}}{x}}  = \sqrt {36}  = 6\\
7,\\
a,\\
\sqrt 3 .x = \sqrt {27} \\
 \Leftrightarrow \sqrt 3 .x = \sqrt {{{3.3}^2}} \\
 \Leftrightarrow \sqrt 3 .x = \sqrt 3 .3\\
 \Leftrightarrow x = 3\\
b,\\
\sqrt 3 .x - \sqrt {27}  = \sqrt {12}  - \sqrt {75} \\
 \Leftrightarrow \sqrt 3 .x - \sqrt {{3^2}.3}  = \sqrt {{2^2}.3}  - \sqrt {{5^2}.3} \\
 \Leftrightarrow \sqrt 3 .x - 3\sqrt 3  = 2\sqrt 3  - 5\sqrt 3 \\
 \Leftrightarrow \sqrt 3 .x - 3\sqrt 3  =  - 3\sqrt 3 \\
 \Leftrightarrow \sqrt 3 .x = 0\\
 \Leftrightarrow x = 0\\
c,\\
\sqrt 5 .{x^2} - \sqrt {20}  = 0\\
 \Leftrightarrow \sqrt 5 .{x^2} - \sqrt {{2^2}.5}  = 0\\
 \Leftrightarrow \sqrt 5 .{x^2} - 2.\sqrt 5  = 0\\
 \Leftrightarrow \sqrt 5 .\left( {{x^2} - 2} \right) = 0\\
 \Leftrightarrow {x^2} = 2\\
 \Leftrightarrow x =  \pm \sqrt 2 \\
d,\\
\dfrac{{2{x^2}}}{{\sqrt 3 }} - \sqrt {12}  = 0\\
 \Leftrightarrow \dfrac{{2{x^2}}}{{\sqrt 3 }} = \sqrt {12} \\
 \Leftrightarrow 2{x^2} = \sqrt 3 .\sqrt {12} \\
 \Leftrightarrow 2{x^2} = \sqrt {36} \\
 \Leftrightarrow 2{x^2} = 6\\
 \Leftrightarrow {x^2} = 3\\
 \Leftrightarrow x =  \pm \sqrt 3 
\end{array}\)