Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\dfrac{{\sqrt {63{y^3}} }}{{\sqrt {7y} }} = \sqrt {\dfrac{{63{y^3}}}{{7y}}}  = \sqrt {9{y^2}}  = \sqrt {{{\left( {3y} \right)}^2}}  = \left| {3y} \right| = 3y\,\,\,\,\,\,\left( {y > 0} \right)\\
2,\\
\dfrac{{\sqrt {48{x^3}} }}{{\sqrt {3{x^5}} }} = \sqrt {\dfrac{{48{x^3}}}{{3{x^5}}}}  = \sqrt {\dfrac{{16}}{{{x^2}}}}  = \sqrt {{{\left( {\dfrac{4}{x}} \right)}^2}}  = \left| {\dfrac{4}{x}} \right| = \dfrac{4}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x > 0} \right)\\
3,\\
\dfrac{{\sqrt {45m{n^2}} }}{{\sqrt {20m} }} = \sqrt {\dfrac{{45m{n^2}}}{{20m}}}  = \sqrt {\dfrac{{9{n^2}}}{4}}  = \sqrt {{{\left( {\dfrac{3}{2}n} \right)}^2}}  = \left| {\dfrac{3}{2}n} \right| = \dfrac{3}{2}n\,\,\,\,\,\,\,\,\left( {m,n > 0} \right)\\
4,\\
\dfrac{{\sqrt {16{a^4}{b^6}} }}{{\sqrt {128{a^6}{b^6}} }} = \sqrt {\dfrac{{16{a^4}{b^6}}}{{128{a^6}{b^6}}}}  = \sqrt {\dfrac{1}{{8{a^2}}}}  = \sqrt {{{\left( {\dfrac{1}{{2\sqrt 2 a}}} \right)}^2}}  = \left| {\dfrac{1}{{2\sqrt 2 a}}} \right| =  - \dfrac{1}{{2\sqrt 2 a}}\,\,\,\,\,\left( {a < 0} \right)\\
5,\\
5xy\sqrt {\dfrac{{25{x^2}}}{{{y^6}}}}  = 5xy\sqrt {{{\left( {\dfrac{{5x}}{{{y^3}}}} \right)}^2}}  = 5xy.\left| {\dfrac{{5x}}{{{y^3}}}} \right| = 5xy.\dfrac{{ - 5x}}{{{y^3}}} = \dfrac{{ - 25{x^2}}}{{{y^2}}}\,\,\,\,\,\,\,\,\left( {x < 0,\,\,y > 0} \right)\\
6,\\
0,2{x^3}{y^3}.\sqrt {\dfrac{{16}}{{{x^4}{y^8}}}}  = 0,2.{x^3}{y^3}.\sqrt {{{\left( {\dfrac{4}{{{x^2}{y^4}}}} \right)}^2}}  = 0,2.{x^3}{y^3}.\dfrac{4}{{{x^2}{y^4}}} = \dfrac{{0,8x}}{y}\\
7,\\
a{b^2}\sqrt {\dfrac{3}{{{a^2}{b^4}}}}  = a{b^2}.\sqrt {{{\left( {\dfrac{{\sqrt 3 }}{{a{b^2}}}} \right)}^2}}  = a{b^2}.\left| {\dfrac{{\sqrt 3 }}{{a{b^2}}}} \right| = a{b^2}.\dfrac{{ - \sqrt 3 }}{{a{b^2}}} =  - \sqrt 3 \,\,\,\,\left( {a < 0} \right)\\
8,\\
\sqrt {\dfrac{{27{{\left( {a - 3} \right)}^2}}}{{48}}}  = \sqrt {\dfrac{{9{{\left( {a - 3} \right)}^2}}}{{16}}}  = \sqrt {{{\left( {\dfrac{{3\left( {a - 3} \right)}}{4}} \right)}^2}}  = \left| {\dfrac{{3\left( {a - 3} \right)}}{4}} \right| = \dfrac{{3\left( {a - 3} \right)}}{4}\,\,\,\,\,\left( {a > 3 \Rightarrow a - 3 > 0} \right)
\end{array}\)

\(\begin{array}{l}
9,\\
\sqrt {\dfrac{{9 + 12a + 4{a^2}}}{{{b^2}}}}  = \sqrt {\dfrac{{{{\left( {2a + 3} \right)}^2}}}{{{b^2}}}}  = \left| {\dfrac{{2a + 3}}{b}} \right| =  - \dfrac{{2a + 3}}{b}\\
\left( {a \ge  - 1,5 \Rightarrow 2a + 3 \ge 0,\,\,\,b < 0 \Rightarrow \dfrac{{2a + 3}}{b} < 0} \right)\\
10,\\
\sqrt {\dfrac{{x - 2\sqrt x  + 1}}{{x + 2\sqrt x  + 1}}}  = \sqrt {\dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}}}{{{{\left( {\sqrt x  + 1} \right)}^2}}}}  = \left| {\dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}} \right| = \dfrac{{\left| {\sqrt x  - 1} \right|}}{{\sqrt x  + 1}}\\
TH1:\,\,\,\,0 \le x \le 1 \Rightarrow \sqrt x  - 1 \le 0\\
 \Rightarrow \dfrac{{\left| {\sqrt x  - 1} \right|}}{{\sqrt x  + 1}} = \dfrac{{1 - \sqrt x }}{{\sqrt x  + 1}}\\
TH2:\,\,\,\,x > 1 \Rightarrow \sqrt x  - 1 > 0\\
 \Rightarrow \dfrac{{\left| {\sqrt x  - 1} \right|}}{{\sqrt x  + 1}} = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}\\
11,\\
\left( {a - b} \right).\sqrt {\dfrac{{ab}}{{{{\left( {a - b} \right)}^2}}}}  = \left( {a - b} \right).\left| {\dfrac{{\sqrt {ab} }}{{a - b}}} \right| = \left( {a - b} \right).\dfrac{{\sqrt {ab} }}{{b - a}} =  - \sqrt {ab} \\
\left( {a < b < 0 \Rightarrow \left\{ \begin{array}{l}
ab > 0\\
a - b < 0
\end{array} \right.} \right)\\
12,\\
\dfrac{{x - 1}}{{\sqrt y  - 1}}.\sqrt {\dfrac{{{{\left( {y - 2\sqrt y  + 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^4}}}} \\
 = \dfrac{{\left( {x - 1} \right)}}{{\sqrt y  - 1}}.\sqrt {\dfrac{{{{\left[ {{{\left( {\sqrt y  - 1} \right)}^2}} \right]}^2}}}{{{{\left( {x - 1} \right)}^4}}}} \\
 = \dfrac{{x - 1}}{{\sqrt y  - 1}}.\left| {\dfrac{{{{\left( {\sqrt y  - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}}} \right|\\
 = \dfrac{{x - 1}}{{\sqrt y  - 1}}.\dfrac{{{{\left( {\sqrt y  - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{{\sqrt y  - 1}}{{x - 1}}
\end{array}\)