Đáp án:
\(\begin{array}{l}
B4:\\
a)2\sqrt 5 \\
c)2\sqrt 7 \\
b)2\sqrt 2 - 4\\
d)2\sqrt {11} - 1\\
B5:\\
a)A = 2\sqrt 3 \\
b)B = 2\sqrt 5 \\
c)C = - 5x\\
d)D = 2x - 8\\
B6:\\
a)\left[ \begin{array}{l}
A = 1 - 4a\\
A = - 1
\end{array} \right.\\
b)\left[ \begin{array}{l}
B = 0\\
B = 2x - 4y
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B4:\\
a)\left| {\sqrt 5 - \sqrt 2 } \right| + \left| {\sqrt 5 + \sqrt 2 } \right|\\
= \sqrt 5 - \sqrt 2 + \sqrt 5 + \sqrt 2 = 2\sqrt 5 \\
c)\left| {\sqrt 7 - \sqrt 2 } \right| + \left| {\sqrt 7 + \sqrt 2 } \right|\\
= \sqrt 7 - \sqrt 2 + \sqrt 7 + \sqrt 2 \\
= 2\sqrt 7 \\
b)\left| {\sqrt 2 + 1} \right| - \left| {\sqrt 2 - 5} \right|\\
= \sqrt 2 + 1 - \left( {5 - \sqrt 2 } \right)\\
= \sqrt 2 + 1 - 5 + \sqrt 2 \\
= 2\sqrt 2 - 4\\
d)\left| {\sqrt {11} + 4} \right| - \left| {\sqrt {11} - 5} \right|\\
= \sqrt {11} + 4 - \left( {5 - \sqrt {11} } \right)\\
= 2\sqrt {11} - 1\\
B5:\\
a)A = \sqrt {3 + 2.\sqrt 3 .1 + 1} + \sqrt {3 - 2.\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= \sqrt 3 + 1 + \sqrt 3 - 1 = 2\sqrt 3 \\
b)B = \sqrt {5 + 2.\sqrt 5 .1 + 1} + \sqrt {5 - 2.\sqrt 5 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \sqrt 5 + 1 + \sqrt 5 - 1 = 2\sqrt 5 \\
c)C = 3\left| x \right| - 2x = - 3x - 2x = - 5x\\
d)D = x - 4 + \sqrt {{{\left( {4 - x} \right)}^2}} \\
= x - 4 + \left| {4 - x} \right|\\
= x - 4 - 4 + x = 2x - 8\\
B6:\\
a)A = \sqrt {{{\left( {1 - 2a} \right)}^2}} - 2a\\
= \left| {1 - 2a} \right| - 2a\\
\to \left[ \begin{array}{l}
A = 1 - 2a - 2a\left( {DK:\dfrac{1}{2} \ge a} \right)\\
A = - 1 + 2a - 2a\left( {DK:\dfrac{1}{2} < a} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = 1 - 4a\\
A = - 1
\end{array} \right.\\
b)B = x - 2y - \sqrt {{{\left( {x - 2y} \right)}^2}} \\
= x - 2y - \left| {x - 2y} \right|\\
\to \left[ \begin{array}{l}
B = x - 2y - x + 2y\left( {DK:x \ge 2y} \right)\\
B = x - 2y + x - 2y\left( {DK:x < 2y} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
B = 0\\
B = 2x - 4y
\end{array} \right.
\end{array}\)
