Đáp án:
B2:
c. \(\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{3}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a.{\left( {2 - \sqrt 2 } \right)^2} + \sqrt {32} \\
= 4 - 4\sqrt 2 + 2 + 4\sqrt 2 \\
= 6\\
b.\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + \sqrt 3 \\
= 2 - \sqrt 3 + \sqrt 3 = 2\left( {do:2 > \sqrt 3 } \right)\\
B2:\\
a.\sqrt {{{\left( {x - 2} \right)}^2}} - 1 = 5\\
\to \sqrt {{{\left( {x - 2} \right)}^2}} = 6\\
\to \left| {x - 2} \right| = 6\\
\to \left[ \begin{array}{l}
x - 2 = 6\\
x - 2 = - 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 8\\
x = - 4
\end{array} \right.\\
b.\sqrt {9\left( {x - 2} \right)} = 15\\
\to 3\left| {x - 2} \right| = 15\\
\to \left| {x - 2} \right| = 5\\
\to \left[ \begin{array}{l}
x - 2 = 5\\
x - 2 = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = - 3
\end{array} \right.\\
c.\sqrt {{{\left( {2x + 1} \right)}^2}} = 2\\
\to \left| {2x + 1} \right| = 2\\
\to \left[ \begin{array}{l}
2x + 1 = 2\\
2x + 1 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{3}{2}
\end{array} \right.
\end{array}\)
