Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,\,\,10\\
b,\,\,\,\, - 2\\
c,\,\,\,\,11\\
2,\\
a,\\
x = 6\\
b,\\
\left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
3,\\
a,\\
\left( {\sqrt x  - 1} \right).\left( {y\sqrt x  + 1} \right)\\
b,\\
\left( {\sqrt y  + 1} \right)\left( {\sqrt x  - 1} \right)\\
4,\\
a,\\
\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
b,\\
P = \sqrt x  - 1\\
c,\\
0 < x < 1
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
\left( {4\sqrt 2  + 2\sqrt 5 } \right).\sqrt 5  - \sqrt {160} \\
 = 4\sqrt 2 .\sqrt 5  + 2{\sqrt 5 ^2} - \sqrt {16.10} \\
 = 4.\sqrt {2.5}  + 2.5 - \sqrt {{4^2}.10} \\
 = 4\sqrt {10}  + 10 - 4\sqrt {10} \\
 = 10\\
b,\\
\left( {\dfrac{{\sqrt 3  - \sqrt 6 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15}  - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 3  + \sqrt 5 }}\\
 = \left( {\dfrac{{\sqrt 3  - \sqrt 3 .\sqrt 2 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 .\sqrt 3  - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 3  + \sqrt 5 }}\\
 = \left( {\dfrac{{\sqrt 3 .\left( {1 - \sqrt 2 } \right)}}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 .\left( {\sqrt 3  - 1} \right)}}{{1 - \sqrt 3 }}} \right).\left( {\sqrt 3  + \sqrt 5 } \right)\\
 = \left( {\sqrt 3  - \sqrt 5 } \right).\left( {\sqrt 3  + \sqrt 5 } \right)\\
 = {\sqrt 3 ^2} - {\sqrt 5 ^2}\\
 = 3 - 5\\
 =  - 2\\
c,\\
{\left( {\sqrt 6  + \sqrt 5 } \right)^2} - \sqrt {120} \\
 = {\sqrt 6 ^2} + 2.\sqrt 6 .\sqrt 5  + {\sqrt 5 ^2} - \sqrt {4.30} \\
 = 6 + 2\sqrt {30}  + 5 - \sqrt {{2^2}.30} \\
 = 11 + 2\sqrt {30}  - 2\sqrt {30} \\
 = 11\\
2,\\
a,\\
\sqrt {4x - 8}  - 4\sqrt {x - 2}  = 2 - \sqrt {9x - 18} \\
 \Leftrightarrow \sqrt {4.\left( {x - 2} \right)}  - 4\sqrt {x - 2}  = 2 - \sqrt {9\left( {x - 2} \right)} \\
 \Leftrightarrow \sqrt {{2^2}.\left( {x - 2} \right)}  - 4\sqrt {x - 2}  = 2 - \sqrt {{3^2}.\left( {x - 2} \right)} \\
 \Leftrightarrow 2\sqrt {x - 2}  - 4\sqrt {x - 2}  = 2 - 3\sqrt {x - 2} \\
 \Leftrightarrow 2\sqrt {x - 2}  - 4\sqrt {x - 2}  + 3\sqrt {x - 2}  = 2\\
 \Leftrightarrow \sqrt {x - 2}  = 2\\
 \Leftrightarrow x - 2 = {2^2}\\
 \Leftrightarrow x - 2 = 4\\
 \Leftrightarrow x = 6\\
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,\\
xy - y\sqrt x  + \sqrt x  - 1\\
 = \left( {xy - y\sqrt x } \right) + \left( {\sqrt x  - 1} \right)\\
 = y.\left( {x - \sqrt x } \right) + \left( {\sqrt x  - 1} \right)\\
 = y.\left( {{{\sqrt x }^2} - \sqrt x } \right) + \left( {\sqrt x  - 1} \right)\\
 = y.\sqrt x .\left( {\sqrt x  - 1} \right) + \left( {\sqrt x  - 1} \right)\\
 = \left( {\sqrt x  - 1} \right).\left( {y\sqrt x  + 1} \right)\\
b,\\
\sqrt {xy}  + \sqrt x  - \sqrt y  - 1\\
 = \left( {\sqrt {xy}  + \sqrt x } \right) - \left( {\sqrt y  + 1} \right)\\
 = \sqrt x .\left( {\sqrt y  + 1} \right) - \left( {\sqrt y  + 1} \right)\\
 = \left( {\sqrt y  + 1} \right)\left( {\sqrt x  - 1} \right)\\
4,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  - 1 \ne 0\\
x + \sqrt x  \ne 0\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  \ne 1\\
x\left( {\sqrt x  + 1} \right) \ne 0\\
x \ne 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
b,\\
P = \left( {\dfrac{{x - \sqrt x }}{{\sqrt x  - 1}} - \dfrac{{\sqrt x  + 1}}{{x + \sqrt x }}} \right):\dfrac{{\sqrt x  + 1}}{x}\\
 = \left( {\dfrac{{{{\sqrt x }^2} - \sqrt x }}{{\sqrt x  - 1}} - \dfrac{{\sqrt x  + 1}}{{{{\sqrt x }^2} + \sqrt x }}} \right):\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \left( {\dfrac{{\sqrt x .\left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x .\left( {\sqrt x  + 1} \right)}}} \right):\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \left( {\sqrt x  - \dfrac{1}{{\sqrt x }}} \right):\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \dfrac{{{{\sqrt x }^2} - 1}}{{\sqrt x }}:\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x  + 1}}\\
 = \sqrt x  - 1\\
c,\\
P < 0\\
 \Leftrightarrow \sqrt x  - 1 < 0\\
 \Leftrightarrow \sqrt x  < 1\\
 \Leftrightarrow x < 1\\
x > 0,x \ne 1 \Rightarrow 0 < x < 1
\end{array}\)