Đáp án:
\(\begin{array}{l}
a,\,\,\,\,\dfrac{{33}}{5}\\
b,\,\,\,\, - 4\\
c,\,\,\,\,\, - \sqrt 6 \\
d,\,\,\,\,\, - \sqrt 5
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {25} + \dfrac{{\sqrt {49} }}{{\sqrt 4 }} - \sqrt {\dfrac{9}{4}} - \sqrt {0,16} \\
= \sqrt {{5^2}} + \dfrac{{\sqrt {{7^2}} }}{{\sqrt {{2^2}} }} - \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2}} - \sqrt {{{0,4}^2}} \\
= 5 + \dfrac{7}{2} - \dfrac{3}{2} - 0,4\\
= \dfrac{{33}}{5}\\
b,\\
\sqrt {{{\left( {2\sqrt 5 - 1} \right)}^2}} - \sqrt {{{\left( {3 + 2\sqrt 5 } \right)}^2}} \\
= \left| {2\sqrt 5 - 1} \right| - \left| {3 + 2\sqrt 5 } \right|\\
= \left( {2\sqrt 5 - 1} \right) - \left( {3 + 2\sqrt 5 } \right)\\
= 2\sqrt 5 - 1 - 3 - 2\sqrt 5 \\
= - 4\\
c,\\
\sqrt {11 + 2\sqrt {30} } - \sqrt {{{\left( {\sqrt 5 + 2\sqrt 6 } \right)}^2}} \\
= \sqrt {6 + 2\sqrt {30} + 5} - \left| {\sqrt 5 + 2\sqrt 6 } \right|\\
= \sqrt {{{\sqrt 6 }^2} + 2.\sqrt 6 .\sqrt 5 + {{\sqrt 5 }^2}} - \left( {\sqrt 5 + 2\sqrt 6 } \right)\\
= \sqrt {{{\left( {\sqrt 6 + \sqrt 5 } \right)}^2}} - \left( {\sqrt 5 + 2\sqrt 6 } \right)\\
= \left| {\sqrt 6 + \sqrt 5 } \right| - \sqrt 5 - 2\sqrt 6 \\
= \sqrt 6 + \sqrt 5 - \sqrt 5 - 2\sqrt 6 \\
= - \sqrt 6 \\
d,\\
\sqrt {8 - 2\sqrt {15} } - \sqrt {23 - 4\sqrt {15} } \\
= \sqrt {5 - 2\sqrt {15} + 3} - \sqrt {20 - 4\sqrt {15} + 3} \\
= \sqrt {{{\sqrt 5 }^2} - 2.\sqrt 5 .\sqrt 3 + {{\sqrt 3 }^2}} - \sqrt {{{\left( {2\sqrt 5 } \right)}^2} - 2.2\sqrt 5 .\sqrt 3 + {{\sqrt 3 }^2}} \\
= \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {2\sqrt 5 - \sqrt 3 } \right)}^2}} \\
= \left| {\sqrt 5 - \sqrt 3 } \right| - \left| {2\sqrt 5 - \sqrt 3 } \right|\\
= \left( {\sqrt 5 - \sqrt 3 } \right) - \left( {2\sqrt 5 - \sqrt 3 } \right)\\
= \sqrt 5 - \sqrt 3 - 2\sqrt 5 + \sqrt 3 \\
= - \sqrt 5
\end{array}\)
