Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,\,\,2\\
b,\,\,\,\,0\\
c,\,\,\,\,10\\
2,\\
a,\\
x = 15\\
b,\\
\left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
3,\\
a,\\
\left( {\sqrt x  - \sqrt y } \right){\left( {\sqrt x  + \sqrt y } \right)^2}\\
b,\\
\sqrt {2 + \sqrt 3 }  + \sqrt {2 - \sqrt 3 }  = \sqrt 6 \\
4,\\
a,\\
\left[ \begin{array}{l}
A =  - 1,\,\,\,\,3 \le x \le 4\\
A = 2\sqrt {x - 3}  - 3,\,\,\,\,4 < x < 7\\
A = 1,\,\,\,\,\,x \ge 7
\end{array} \right.\\
b,\\
A =  - 1
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
{\left( {\sqrt {4 - \sqrt 7 }  - \sqrt {4 + \sqrt 7 } } \right)^2}\\
 = {\sqrt {4 - \sqrt 7 } ^2} - 2.\sqrt {\left( {4 - \sqrt 7 } \right)\left( {4 + \sqrt 7 } \right)}  + {\sqrt {4 + \sqrt 7 } ^2}\\
 = \left( {4 - \sqrt 7 } \right) - 2.\sqrt {{4^2} - {{\sqrt 7 }^2}}  + \left( {4 + \sqrt 7 } \right)\\
 = 4 - \sqrt 7  - 2.\sqrt {16 - 7}  + 4 + \sqrt 7 \\
 = 8 - 2\sqrt 9 \\
 = 8 - 2.3\\
 = 8 - 6\\
 = 2\\
b,\\
\dfrac{{5\sqrt 3  - 3\sqrt 5 }}{{\sqrt {15} }} + \dfrac{{3 - \sqrt 6 }}{{\sqrt 3  - \sqrt 2 }} - \dfrac{{10}}{{2\sqrt 5 }}\\
 = \dfrac{{{{\sqrt 5 }^2}.\sqrt 3  - {{\sqrt 3 }^2}.\sqrt 5 }}{{\sqrt {15} }} + \dfrac{{{{\sqrt 3 }^2} - \sqrt 3 .\sqrt 2 }}{{\sqrt 3  - \sqrt 2 }} - \dfrac{{2.5}}{{2\sqrt 5 }}\\
 = \dfrac{{\sqrt 5 .\sqrt 3 .\left( {\sqrt 5  - \sqrt 3 } \right)}}{{\sqrt {15} }} + \dfrac{{\sqrt 3 .\left( {\sqrt 3  - \sqrt 2 } \right)}}{{\sqrt 3  - \sqrt 2 }} - \dfrac{{2.{{\sqrt 5 }^2}}}{{2.\sqrt 5 }}\\
 = \dfrac{{\sqrt {15} .\left( {\sqrt 5  - \sqrt 3 } \right)}}{{\sqrt {15} }} + \sqrt 3  - \sqrt 5 \\
 = \sqrt 5  - \sqrt 3  + \sqrt 3  - \sqrt 5 \\
 = 0\\
c,\\
\left( {5\sqrt 2  + 2\sqrt 5 } \right)\sqrt 5  - \sqrt {250} \\
 = 5\sqrt 2 .\sqrt 5  + 2.{\sqrt 5 ^2} - \sqrt {25.10} \\
 = 5.\sqrt {2.5}  + 2.5 - \sqrt {{5^2}.10} \\
 = 5\sqrt {10}  + 10 - 5\sqrt {10} \\
 = 10\\
2,\\
a,\\
\sqrt {16x + 16}  + \sqrt {4x + 4}  = 16 - \sqrt {x + 1}  + \sqrt {9x + 9} \\
 \Leftrightarrow \sqrt {16.\left( {x + 1} \right)}  + \sqrt {4.\left( {x + 1} \right)}  = 16 - \sqrt {x + 1}  + \sqrt {9\left( {x + 1} \right)} \\
 \Leftrightarrow \sqrt {{4^2}.\left( {x + 1} \right)}  + \sqrt {{2^2}.\left( {x + 1} \right)}  = 16 - \sqrt {x + 1}  + \sqrt {{3^2}.\left( {x + 1} \right)} \\
 \Leftrightarrow 4\sqrt {x + 1}  + 2\sqrt {x + 1}  = 16 - \sqrt {x + 1}  + 3\sqrt {x + 1} \\
 \Leftrightarrow 6\sqrt {x + 1}  = 16 + 2\sqrt {x + 1} \\
 \Leftrightarrow 4\sqrt {x + 1}  = 16\\
 \Leftrightarrow \sqrt {x + 1}  = 4\\
 \Leftrightarrow x + 1 = {4^2}\\
 \Leftrightarrow x + 1 = 16\\
 \Leftrightarrow x = 15\\
b,\\
\sqrt {4{x^2} - 4x + 1}  = 3\\
 \Leftrightarrow \sqrt {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}}  = 3\\
 \Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}}  = 3\\
 \Leftrightarrow \left| {2x - 1} \right| = 3\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 3\\
2x - 1 =  - 3
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 4\\
2x =  - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
3,\\
a,\\
x,y \ge 0 \Rightarrow \left| x \right| = x;\,\,\left| y \right| = y\\
\sqrt {{x^3}}  - \sqrt {{y^3}}  + \sqrt {{x^2}y}  - \sqrt {x{y^2}} \\
 = \sqrt {{x^2}.x}  - \sqrt {{y^2}.y}  + \sqrt {{x^2}.y}  - \sqrt {{y^2}.x} \\
 = \left| x \right|.\sqrt x  - \left| y \right|.\sqrt y  + \left| x \right|.\sqrt y  - \left| y \right|.\sqrt x \\
 = x\sqrt x  - y\sqrt y  + x\sqrt y  - y\sqrt x \\
 = \left( {x\sqrt x  - y\sqrt x } \right) + \left( {x\sqrt y  - y\sqrt y } \right)\\
 = \sqrt x .\left( {x - y} \right) + \sqrt y .\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {\sqrt x  + \sqrt y } \right)\\
 = \left( {{{\sqrt x }^2} - {{\sqrt y }^2}} \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 } \right)^2}\\
b,\\
\sqrt {2 + \sqrt 3 }  + \sqrt {2 - \sqrt 3 } \\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 2 .\sqrt {2 + \sqrt 3 }  + \sqrt 2 .\sqrt {2 - \sqrt 3 } } \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {2.\left( {2 + \sqrt 3 } \right)}  + \sqrt {2.\left( {2 - \sqrt 3 } \right)} } \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {4 + 2\sqrt 3 }  + \sqrt {4 - 2\sqrt 3 } } \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {3 + 2\sqrt 3  + 1}  + \sqrt {3 - 2\sqrt 3  + 1} } \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\sqrt 3 }^2} + 2.\sqrt 3 .1 + {1^2}}  + \sqrt {{{\sqrt 3 }^2} - 2.\sqrt 3 .1 + {1^2}} } \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}}  + \sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}} } \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 3  + 1 + \sqrt 3  - 1} \right)\\
 = \dfrac{1}{{\sqrt 2 }}.2\sqrt 3 \\
 = \sqrt 2 .\sqrt 3 \\
 = \sqrt 6 \\
4,\\
a,\\
DKXD:\,\,\,\,x \ge 3\\
A = \sqrt {x - 2 - 2\sqrt {x - 3} }  - \sqrt {x + 1 - 4\sqrt {x - 3} } \\
 = \sqrt {\left( {x - 3} \right) - 2\sqrt {x - 3}  + 1}  - \sqrt {\left( {x - 3} \right) - 4\sqrt {x - 3}  + 4} \\
 = \sqrt {{{\sqrt {x - 3} }^2} - 2.\sqrt {x - 3} .1 + {1^2}}  - \sqrt {{{\sqrt {x - 3} }^2} - 2.\sqrt {x - 3} .2 + {2^2}} \\
 = \sqrt {{{\left( {\sqrt {x - 3}  - 1} \right)}^2}}  - \sqrt {{{\left( {\sqrt {x - 3}  - 2} \right)}^2}} \\
 = \left| {\sqrt {x - 3}  - 1} \right| - \left| {\sqrt {x - 3}  - 2} \right|\\
TH1:\,\,\,\sqrt {x - 3}  \le 1 \Leftrightarrow x - 3 \le 1 \Leftrightarrow 3 \le x \le 4\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt {x - 3}  - 1 \le 0\\
\sqrt {x - 3}  - 2 \le 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {\sqrt {x - 3}  - 1} \right| =  - \left( {\sqrt {x - 3}  - 1} \right)\\
\left| {\sqrt {x - 3}  - 2} \right| =  - \left( {\sqrt {x - 3}  - 2} \right)
\end{array} \right.\\
A = \left| {\sqrt {x - 3}  - 1} \right| - \left| {\sqrt {x - 3}  - 2} \right|\\
 =  - \left( {\sqrt {x - 3}  - 1} \right) + \left( {\sqrt {x - 3}  - 2} \right)\\
 =  - \sqrt {x - 3}  + 1 + \sqrt {x - 3}  - 2\\
 =  - 1\\
TH2:\,\,\,1 < \sqrt {x - 3}  < 2 \Leftrightarrow 1 < x - 3 < 4 \Leftrightarrow 4 < x < 7\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt {x - 3}  - 1 > 0\\
\sqrt {x - 3}  - 2 < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {\sqrt {x - 3}  - 1} \right| = \sqrt {x - 3}  - 1\\
\left| {\sqrt {x - 3}  - 2} \right| =  - \left( {\sqrt {x - 3}  - 2} \right)
\end{array} \right.\\
A = \left| {\sqrt {x - 3}  - 1} \right| - \left| {\sqrt {x - 3}  - 2} \right|\\
 = \left( {\sqrt {x - 3}  - 1} \right) + \left( {\sqrt {x - 3}  - 2} \right)\\
 = 2\sqrt {x - 3}  - 3\\
TH3:\,\,\,\sqrt {x - 3}  \ge 2 \Rightarrow x - 3 \ge 4 \Leftrightarrow x \ge 7\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt {x - 3}  - 1 > 0\\
\sqrt {x - 3}  - 2 \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {\sqrt {x - 3}  - 1} \right| = \sqrt {x - 3}  - 1\\
\left| {\sqrt {x - 3}  - 2} \right| = \sqrt {x - 3}  - 2
\end{array} \right.\\
A = \left| {\sqrt {x - 3}  - 1} \right| - \left| {\sqrt {x - 3}  - 2} \right|\\
 = \left( {\sqrt {x - 3}  - 1} \right) - \left( {\sqrt {x - 3}  - 2} \right)\\
 = \sqrt {x - 3}  - 1 - \sqrt {x - 3}  + 2\\
 = 1\\
 \Rightarrow \left[ \begin{array}{l}
A =  - 1,\,\,\,\,3 \le x \le 4\\
A = 2\sqrt {x - 3}  - 3,\,\,\,\,4 < x < 7\\
A = 1,\,\,\,\,\,x \ge 7
\end{array} \right.\\
b,\\
3 \le x \le 4 \Rightarrow A =  - 1
\end{array}\)