Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
a,\\
C = \sqrt {16{x^2} + 8x + 1} = \sqrt {{{\left( {4x} \right)}^2} + 2.4x.1 + {1^2}} = \sqrt {{{\left( {4x + 1} \right)}^2}} = \left| {4x + 1} \right|\\
b,\\
x = \sqrt 3 \Rightarrow C = \left| {4.\sqrt 3 + 1} \right| = 4\sqrt 3 + 1\\
4,\\
x = \sqrt 2 \Rightarrow D = \sqrt {3.{{\sqrt 2 }^2} - 2.\sqrt 2 - \sqrt 2 .\sqrt 2 - 1} \\
= \sqrt {3.2 - 2\sqrt 2 - 2 - 1} = \sqrt {3 - 2\sqrt 2 } = \sqrt {2 - 2.\sqrt 2 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} = \left| {\sqrt 2 - 1} \right| = \sqrt 2 - 1\\
5,\\
a,\\
E = \sqrt {1 + 4x + 4{x^2}} = \sqrt {{1^2} + 2.1.2x + {{\left( {2x} \right)}^2}} = \sqrt {{{\left( {1 + 2x} \right)}^2}} = \left| {1 + 2x} \right|\\
b,\\
x = 7 \Rightarrow E = \left| {1 + 2.7} \right| = 15\\
6,\\
a.\\
F = \sqrt {{x^2} - 3.\left( {2x - 3} \right)} = \sqrt {{x^2} - 6x + 9} = \sqrt {{x^2} - 2.x.3 + {3^2}} = \sqrt {{{\left( {x - 3} \right)}^2}} = \left| {x - 3} \right|\\
b,\\
x = 8 \Rightarrow F = \left| {8 - 3} \right| = 5
\end{array}\)
