Đáp án:
\(\begin{array}{l}
B1:\\
a)x = 7\\
b)x = \dfrac{{18}}{5}\\
c)\left[ \begin{array}{l}
x = 4\\
x = \dfrac{{10}}{3}
\end{array} \right.\\
B2:\\
a)\dfrac{{8\sqrt 5 }}{{15}}\\
b)\dfrac{{3\sqrt 2 }}{2}\\
c)\dfrac{{\sqrt 2 + 2}}{6}\\
d)\sqrt 5 + 2\\
e)\sqrt 5 - 1\\
f)\dfrac{{\sqrt {10} - \sqrt 6 }}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:x \ge 3\\
\sqrt {4\left( {x - 3} \right)} - 2.\dfrac{{\sqrt {x - 3} }}{2} = 2\\
\to 2\sqrt {x - 3} - \sqrt {x - 3} = 2\\
\to \sqrt {x - 3} = 2\\
\to x - 3 = 4\\
\to x = 7\\
b)DK:x \ge \dfrac{2}{5}\\
\sqrt {9\left( {5x - 2} \right)} + \sqrt {4\left( {5x - 2} \right)} = 20\\
\to 3\sqrt {5x - 2} + 2\sqrt {5x - 2} = 20\\
\to \sqrt {5x - 2} = 4\\
\to 5x - 2 = 16\\
\to 5x = 18\\
\to x = \dfrac{{18}}{5}\\
c)\sqrt {{x^2} - 6x + 9} = 2x - 7\\
\to \sqrt {{{\left( {x - 3} \right)}^2}} = 2x - 7\\
\to \left| {x - 3} \right| = 2x - 7\\
\to \left[ \begin{array}{l}
x - 3 = 2x - 7\left( {DK:x \ge 3} \right)\\
x - 3 = - 2x + 7\left( {DK:x < 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
3x = 10
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = \dfrac{{10}}{3}
\end{array} \right.\\
B2:\\
a)\dfrac{8}{{3\sqrt 5 }} = \dfrac{{8\sqrt 5 }}{{15}}\\
b)\sqrt {\dfrac{9}{2}} = \dfrac{3}{{\sqrt 2 }} = \dfrac{{3\sqrt 2 }}{2}\\
c)\dfrac{{1 + \sqrt 2 }}{{3\sqrt 2 }} = \dfrac{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}}{6}\\
= \dfrac{{\sqrt 2 + 2}}{6}\\
d)\dfrac{1}{{\sqrt 5 - 2}} = \dfrac{{\sqrt 5 + 2}}{{5 - 4}} = \sqrt 5 + 2\\
e)\dfrac{4}{{\sqrt 5 + 1}} = \dfrac{{4\left( {\sqrt 5 - 1} \right)}}{{5 - 1}} = \sqrt 5 - 1\\
f)\dfrac{2}{{\sqrt {10} + \sqrt 6 }} = \dfrac{{2\left( {\sqrt {10} - \sqrt 6 } \right)}}{{10 - 6}}\\
= \dfrac{{2\left( {\sqrt {10} - \sqrt 6 } \right)}}{4} = \dfrac{{\sqrt {10} - \sqrt 6 }}{2}
\end{array}\)
