Đáp án:
B2:
d. 2
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
1)a.\sqrt {{{89}^2}} = 89\\
b.\sqrt {2017} + \left| {\sqrt {2017} - 2016} \right|\\
= \sqrt {2017} + 2016 - \sqrt {2017} \left( {do:2016 > \sqrt {2017} } \right)\\
= 2016\\
2)a.DK: - 2x + 3 \ge 0\\
\to \dfrac{3}{2} \ge x\\
b.DK:x \ne 0\\
B2:\\
a.5\sqrt 3 + 4\sqrt 3 - 10\sqrt 3 \\
= - \sqrt 3 \\
b.4\sqrt a + 2.2\sqrt {10a} - 3.3\sqrt {10a} \\
= 4\sqrt a - 5\sqrt {10a} \\
c.\left| {2\sqrt 6 - 3} \right| + \sqrt {9 - 2.3.\sqrt 6 + 6} \\
= 2\sqrt 6 - 3 + \sqrt {{{\left( {3 - \sqrt 6 } \right)}^2}} \left( {Do:2\sqrt 6 > 3} \right)\\
= 2\sqrt 6 - 3 + 3 - \sqrt 6 \left( {do:\sqrt 6 < 3} \right)\\
= \sqrt 6 \\
d.\dfrac{{2\left( {\sqrt 3 + 1} \right) - 2\left( {\sqrt 3 - 1} \right)}}{{3 - 1}}\\
= \dfrac{{2\sqrt 3 + 2 - 2\sqrt 3 + 2}}{2} = 2
\end{array}\)
