Đáp án:
\(\begin{array}{l}
1.\\
1,\\
\dfrac{{17}}{3}\\
2,\\
2\sqrt[3]{5} > \dfrac{1}{2}\sqrt[3]{{311}}\\
3,\\
- \dfrac{{3\sqrt 5 + 7}}{4}\\
2.\\
1,\\
a \le 3\\
2,\\
P = - 5a + 5
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1.\\
1,\\
\sqrt 3 .\left( {\sqrt {12} - \sqrt {\dfrac{1}{{27}}} } \right)\\
= \sqrt 3 .\left( {\sqrt {4.3} - \sqrt {\dfrac{1}{{81}}.3} } \right)\\
= \sqrt 3 .\left( {\sqrt {{2^2}.3} - \sqrt {{{\left( {\dfrac{1}{9}} \right)}^2}.3} } \right)\\
= \sqrt 3 .\left( {2\sqrt 3 - \dfrac{1}{9}\sqrt 3 } \right)\\
= \sqrt 3 .\sqrt 3 .\left( {2 - \dfrac{1}{9}} \right)\\
= {\sqrt 3 ^2}.\dfrac{{17}}{9}\\
= 3.\dfrac{{17}}{9}\\
= \dfrac{{17}}{3}\\
2,\\
2\sqrt[3]{5} = \sqrt[3]{{{2^3}.5}} = \sqrt[3]{{8.5}} = \sqrt[3]{{40}}\\
\dfrac{1}{2}\sqrt[3]{{311}} = \sqrt[3]{{{{\left( {\dfrac{1}{2}} \right)}^3}.311}} = \sqrt[3]{{\dfrac{1}{8}.311}} < \sqrt[3]{{\dfrac{1}{8}.320}} = \sqrt[3]{{40}} = 2\sqrt[3]{5}\\
\Rightarrow 2\sqrt[3]{5} > \dfrac{1}{2}\sqrt[3]{{311}}\\
3,\\
\dfrac{1}{{3\sqrt 5 - 7}} = \dfrac{{3\sqrt 5 + 7}}{{\left( {3\sqrt 5 - 7} \right)\left( {3\sqrt 5 + 7} \right)}}\\
= \dfrac{{3\sqrt 5 + 7}}{{{{\left( {3\sqrt 5 } \right)}^2} - {7^2}}} = \dfrac{{3\sqrt 5 + 7}}{{{3^2}.{{\sqrt 5 }^2} - 49}}\\
= \dfrac{{3\sqrt 5 + 7}}{{9.5 - 49}} = \dfrac{{3\sqrt 5 + 7}}{{45 - 49}} = \dfrac{{3\sqrt 5 + 7}}{{ - 4}} = - \dfrac{{3\sqrt 5 + 7}}{4}\\
2.\\
1,\\
\sqrt {9 - 3a} \,\,có\,\,nghĩa\,\,khi\,\,và\,\,chỉ\,\,khi:\\
9 - 3a \ge 0 \Leftrightarrow 9 \ge 3a \Leftrightarrow 3a \le 9 \Leftrightarrow a \le 3\\
2,\\
a \le 1 \Rightarrow a - 1 \le 0 \Rightarrow \left| {a - 1} \right| = - \left( {a - 1} \right)\\
P = \sqrt {\dfrac{{15}}{2}} .\sqrt {\dfrac{{10{{\left( {a - 1} \right)}^2}}}{3}} \\
= \sqrt {\dfrac{{15}}{2}.\dfrac{{10{{\left( {a - 1} \right)}^2}}}{3}} \\
= \sqrt {\dfrac{{15.10}}{{2.3}}.{{\left( {a - 1} \right)}^2}} \\
= \sqrt {25{{\left( {a - 1} \right)}^2}} \\
= \sqrt {{5^2}.{{\left( {a - 1} \right)}^2}} \\
= \left| {5.\left( {a - 1} \right)} \right|\\
= 5.\left| {a - 1} \right|\\
= - 5.\left( {a - 1} \right)\\
= - 5a + 5
\end{array}\)
