Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\sqrt {36} - \sqrt {64} = \sqrt {{6^2}} - \sqrt {{8^2}} = 6 - 8 = - 2\\
b,\\
2\sqrt {12} + 3\sqrt {27} - \sqrt {48} \\
= 2\sqrt {4.3} + 3\sqrt {9.3} - \sqrt {16.3} \\
= 2\sqrt {{2^2}.3} + 3\sqrt {{3^2}.3} - \sqrt {{4^2}.3} \\
= 2.2.\sqrt 3 + 3.3.\sqrt 3 - 4\sqrt 3 \\
= 4\sqrt 3 + 9\sqrt 3 - 4\sqrt 3 \\
= 9\sqrt 3 \\
c,\\
\left( {\sqrt {28} - \sqrt {12} + \sqrt 7 } \right).\sqrt 7 + 2.\sqrt {21} \\
= \left( {\sqrt {{2^2}.7} - \sqrt {{2^2}.3} + \sqrt 7 } \right).\sqrt 7 + 2\sqrt {21} \\
= \left( {2\sqrt 7 - 2\sqrt 3 + \sqrt 7 } \right).\sqrt 7 + 2\sqrt {21} \\
= \left( {3\sqrt 7 - 2\sqrt 3 } \right).\sqrt 7 + 2\sqrt {21} \\
= 3{\sqrt 7 ^2} - 2.\sqrt 3 .\sqrt 7 + 2\sqrt {21} \\
= 3.7 - 2\sqrt {21} + 2\sqrt {21} \\
= 21\\
d,\\
\sqrt {{{\left( {\sqrt 2 - 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \\
= \left| {\sqrt 2 - 2} \right| - \left| {\sqrt 2 - 1} \right|\\
= \left( {2 - \sqrt 2 } \right) - \left( {\sqrt 2 - 1} \right)\\
= 2 - \sqrt 2 - \sqrt 2 + 1\\
= 3 - 2\sqrt 2 \\
2,\\
a,\\
\sqrt {{{\left( {3x + 1} \right)}^2}} = 5\\
\Leftrightarrow \left| {3x + 1} \right| = 5\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 1 = 5\\
3x + 1 = - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = 4\\
3x = - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = - 2
\end{array} \right.
\end{array}\)
