Đáp án:
\(\begin{array}{l}
1,\\
a,\\
x = 7\\
b,\\
x = \dfrac{{18}}{5}\\
c,\\
x = 4\\
2,\\
a,\\
\dfrac{{8\sqrt 5 }}{{15}}\\
b,\\
\dfrac{{3\sqrt 2 }}{2}\\
c,\\
\dfrac{{\sqrt 2 + 2}}{6}\\
d,\\
\sqrt 5 + 2\\
e,\\
\sqrt 5 - 1\\
f,\\
\dfrac{{\sqrt {10} - \sqrt 6 }}{2}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
DKXD:\,\,\,x \ge 3\\
\sqrt {4x - 12} - 2\sqrt {\dfrac{{x - 3}}{4}} = 2\\
\Leftrightarrow \sqrt {4.\left( {x - 3} \right)} - 2\sqrt {\dfrac{1}{4}.\left( {x - 3} \right)} = 2\\
\Leftrightarrow \sqrt {{2^2}.\left( {x - 3} \right)} - 2.\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2}.\left( {x - 3} \right)} = 2\\
\Leftrightarrow 2\sqrt {x - 3} - 2.\dfrac{1}{2}\sqrt {x - 3} = 2\\
\Leftrightarrow 2\sqrt {x - 3} - \sqrt {x - 3} = 2\\
\Leftrightarrow \sqrt {x - 3} = 2\\
\Leftrightarrow x - 3 = {2^2}\\
\Leftrightarrow x - 3 = 4\\
\Leftrightarrow x = 7\\
b,\\
DKXD:\,\,\,x \ge \dfrac{2}{5}\\
\sqrt {9\left( {5x - 2} \right)} + \sqrt {20x - 8} = 20\\
\Leftrightarrow \sqrt {9\left( {5x - 2} \right)} + \sqrt {4.\left( {5x - 2} \right)} = 20\\
\Leftrightarrow \sqrt {{3^2}.\left( {5x - 2} \right)} + \sqrt {{2^2}.\left( {5x - 2} \right)} = 20\\
\Leftrightarrow 3\sqrt {5x - 2} + 2\sqrt {5x - 2} = 20\\
\Leftrightarrow 5\sqrt {5x - 2} = 20\\
\Leftrightarrow \sqrt {5x - 2} = 4\\
\Leftrightarrow 5x - 2 = {4^2}\\
\Leftrightarrow 5x - 2 = 16\\
\Leftrightarrow 5x = 18\\
\Leftrightarrow x = \dfrac{{18}}{5}\\
c,\\
DKXD:\,\,\,{x^2} - 3.\left( {2x - 3} \right) \ge 0\\
\sqrt {{x^2} - 3.\left( {2x - 3} \right)} = 2x - 7\\
\Leftrightarrow \sqrt {{x^2} - 6x + 9} = 2x - 7\\
\Leftrightarrow \sqrt {{x^2} - 2.x.3 + {3^2}} = 2x - 7\\
\Leftrightarrow \sqrt {{{\left( {x - 3} \right)}^2}} = 2x - 7\\
\Leftrightarrow \left| {x - 3} \right| = 2x - 7\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 7 \ge 0\\
\left[ \begin{array}{l}
x - 3 = 2x - 7\\
x - 3 = - 2x + 7
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{7}{2}\\
\left[ \begin{array}{l}
x - 2x = - 7 + 3\\
x + 2x = 7 + 3
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{7}{2}\\
\left[ \begin{array}{l}
- x = - 4\\
3x = 10
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{7}{2}\\
\left[ \begin{array}{l}
x = 4\\
x = \dfrac{{10}}{3}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow x = 4\\
2,\\
a,\\
\dfrac{8}{{3\sqrt 5 }} = \dfrac{{8.\sqrt 5 }}{{3.{{\sqrt 5 }^2}}} = \dfrac{{8\sqrt 5 }}{{3.5}} = \dfrac{{8\sqrt 5 }}{{15}}\\
b,\\
\sqrt {\dfrac{9}{2}} = \sqrt {\dfrac{{{3^2}}}{2}} = \dfrac{3}{{\sqrt 2 }} = \dfrac{{3\sqrt 2 }}{{{{\sqrt 2 }^2}}} = \dfrac{{3\sqrt 2 }}{2}\\
c,\\
\dfrac{{1 + \sqrt 2 }}{{3\sqrt 2 }} = \dfrac{{\sqrt 2 .\left( {1 + \sqrt 2 } \right)}}{{3{{\sqrt 2 }^2}}} = \dfrac{{\sqrt 2 + {{\sqrt 2 }^2}}}{{3.2}} = \dfrac{{\sqrt 2 + 2}}{6}\\
d,\\
\dfrac{1}{{\sqrt 5 - 2}} = \dfrac{{\sqrt 5 + 2}}{{\left( {\sqrt 5 - 2} \right)\left( {\sqrt 5 + 2} \right)}} = \dfrac{{\sqrt 5 + 2}}{{{{\sqrt 5 }^2} - {2^2}}}\\
= \dfrac{{\sqrt 5 + 2}}{{5 - 4}} = \dfrac{{\sqrt 5 + 2}}{1} = \sqrt 5 + 2\\
e,\\
\dfrac{4}{{\sqrt 5 + 1}} = \dfrac{{4\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 + 1} \right)\left( {\sqrt 5 - 1} \right)}} = \dfrac{{4\left( {\sqrt 5 - 1} \right)}}{{{{\sqrt 5 }^2} - {1^2}}}\\
= \dfrac{{4\left( {\sqrt 5 - 1} \right)}}{{5 - 1}} = \dfrac{{4.\left( {\sqrt 5 - 1} \right)}}{4} = \sqrt 5 - 1\\
f,\\
\dfrac{2}{{\sqrt {10} + \sqrt 6 }} = \dfrac{{2.\left( {\sqrt {10} - \sqrt 6 } \right)}}{{\left( {\sqrt {10} + \sqrt 6 } \right).\left( {\sqrt {10} - \sqrt 6 } \right)}}\\
= \dfrac{{2.\left( {\sqrt {10} - \sqrt 6 } \right)}}{{{{\sqrt {10} }^2} - {{\sqrt 6 }^2}}} = \dfrac{{2.\left( {\sqrt {10} - \sqrt 6 } \right)}}{{10 - 6}}\\
= \dfrac{{2.\left( {\sqrt {10} - \sqrt 6 } \right)}}{4} = \dfrac{{\sqrt {10} - \sqrt 6 }}{2}
\end{array}\)
