Đáp án:
2) b. \(\left[ \begin{array}{l}
x = 4\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a.\sqrt 5 + \sqrt 5 - \sqrt {5 - 2\sqrt 5 .1 + 1} + \left( { - 3} \right)\\
= 2\sqrt 5 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} - 3\\
= 2\sqrt 5 - \sqrt 5 + 1 - 3 = \sqrt 5 - 2\\
b.\left[ {\dfrac{{2\left( {\sqrt 3 - 2} \right) - \sqrt 3 - 2}}{{3 - 4}}} \right].\left( {6 + \sqrt 3 } \right)\\
= \left[ {\dfrac{{2\sqrt 3 - 4 - \sqrt 3 - 2}}{{ - 1}}} \right].\left( {6 + \sqrt 3 } \right)\\
= \left( { - \sqrt 3 + 6} \right).\left( {6 + \sqrt 3 } \right)\\
= 36 - 3 = 33\\
2)a.DK:x \ge 3\\
2\sqrt {9\left( {x - 3} \right)} - \dfrac{1}{5}.\sqrt {25\left( {x - 3} \right)} - \dfrac{1}{7}.\sqrt {49.\left( {x - 3} \right)} = 20\\
\to 2.3\sqrt {x - 3} - \dfrac{1}{5}.5\sqrt {x - 3} - \dfrac{1}{7}.7\sqrt {x - 3} = 20\\
\to \left( {6 - 1 - 1} \right)\sqrt {x - 3} = 20\\
\to 4\sqrt {x - 3} = 20\\
\to \sqrt {x - 3} = 5\\
\to x - 3 = 25\\
\to x = 28\\
b.\sqrt {{{\left( {x - 5} \right)}^2}} = 7 - 2x\\
\to \left| {x - 5} \right| = 7 - 2x\\
\to \left[ \begin{array}{l}
x - 5 = 7 - 2x\\
x - 5 = - 7 + 2x
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 12\\
x = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 2
\end{array} \right.
\end{array}\)
