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}\)