Đáp án:
\(\begin{array}{l}
B8:\\
a)2 - \sqrt 3 \\
b) - 3 + \sqrt {10} \\
c) - 2a\\
d)3a - 6\\
e)\left[ \begin{array}{l}
E = x - 3\\
E = - x + 1
\end{array} \right.
\end{array}\)
Bài 9:
điều phải chứng minh
Giải thích các bước giải:
\(\begin{array}{l}
B8:\\
a)\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} = \left| {2 - \sqrt 3 } \right|\\
= 2 - \sqrt 3 \left( {do:2 > \sqrt 3 } \right)\\
b)\sqrt {{{\left( {3 - \sqrt {10} } \right)}^2}} = \left| {3 - \sqrt {10} } \right|\\
= - 3 + \sqrt {10} \left( {do:3 < \sqrt {10} } \right)\\
c)2\sqrt {{a^2}} = 2\left| a \right| = - 2a\\
d)3\left| {a - 2} \right| = 3\left( {a - 2} \right)\\
= 3a - 6\\
e)E = \left| {x - 2} \right| - 1\\
\to \left[ \begin{array}{l}
E = x - 2 - 1\left( {DK:x \ge 2} \right)\\
E = - x + 2 - 1\left( {DK:x < 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
E = x - 3\\
E = - x + 1
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
B9:\\
a)VT = \sqrt {5 - 2.2.\sqrt 5 + 4} - \sqrt 5 \\
= \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - \sqrt 5 \\
= \sqrt 5 - 2 - \sqrt 5 = - 2 = VP\\
b)VT = 16 - 2.4.\sqrt 7 + 7 = 23 - 8\sqrt 7 = VP\\
c)VT = \sqrt {10 - 2.\sqrt {10} .1 + 1} - \sqrt {10} \\
= \sqrt {{{\left( {\sqrt {10} - 1} \right)}^2}} - \sqrt {10} \\
= \sqrt {10} - 1 - \sqrt {10} = - 1 = VP\\
d)VT = \sqrt {3 + 2\sqrt 3 .1 + 1} - \sqrt {3 - 2\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= \sqrt 3 + 1 - \sqrt 3 + 1 = 2 = VP
\end{array}\)
