Đáp án:
\(\begin{array}{l}
19,\\
a,\,\,\,14\\
b,\,\,\, - \dfrac{{2\sqrt {11} + 3\sqrt 7 }}{{19}}\\
20,\\
a,\,\,\,5{a^2}\\
b,\,\,\,x\\
c,\,\,\, - 1\\
d,\,\,\,\left[ \begin{array}{l}
D = 6,\,\,\,\,x \ge - 3\\
D = - 2x,\,\,\,\,x < - 3
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
19,\\
a,\\
\dfrac{1}{{7 + 4\sqrt 3 }} + \dfrac{1}{{7 - 4\sqrt 3 }}\\
= \dfrac{{1.\left( {7 - 4\sqrt 3 } \right) + 1.\left( {7 + 4\sqrt 3 } \right)}}{{\left( {7 + 4\sqrt 3 } \right).\left( {7 - 4\sqrt 3 } \right)}}\\
= \dfrac{{7 - 4\sqrt 3 + 7 + 4\sqrt 3 }}{{{7^2} - {{\left( {4\sqrt 3 } \right)}^2}}}\\
= \dfrac{{14}}{{49 - 48}}\\
= 14\\
b,\\
\dfrac{1}{{2\sqrt {11} - 3\sqrt 7 }}\\
= \dfrac{{1.\left( {2\sqrt {11} + 3\sqrt 7 } \right)}}{{\left( {2\sqrt {11} - 3\sqrt 7 } \right).\left( {2\sqrt {11} + 3\sqrt 7 } \right)}}\\
= \dfrac{{2\sqrt {11} + 3\sqrt 7 }}{{{{\left( {2\sqrt {11} } \right)}^2} - {{\left( {3\sqrt 7 } \right)}^2}}}\\
= \dfrac{{2\sqrt {11} + 3\sqrt 7 }}{{44 - 63}}\\
= \dfrac{{2\sqrt {11} + 3\sqrt 7 }}{{ - 19}}\\
= - \dfrac{{2\sqrt {11} + 3\sqrt 7 }}{{19}}\\
20,\\
a,\\
\sqrt {9{a^4}} + 2{a^2} = \sqrt {{3^2}.{a^4}} + 2{a^2} = \sqrt {{{\left( {3{a^2}} \right)}^2}} + 2{a^2}\\
= \left| {3{a^2}} \right| + 2{a^2} = 3{a^2} + 2{a^2} = 5{a^2}\\
\left( {{a^2} \ge 0,\,\,\forall a \Rightarrow \left| {{a^2}} \right| = {a^2}} \right)\\
b,\\
\sqrt {9{x^2}} - 2x = \sqrt {{{\left( {3x} \right)}^2}} - 2x = \left| {3x} \right| - 2x = 3x - 2x = x\\
\left( {x \ge 0 \Rightarrow \left| x \right| = x} \right)\\
c,\\
\sqrt {4 - 2\sqrt 3 } - \sqrt 3 \\
= \sqrt {3 - 2.\sqrt 3 .1 + 1} - \sqrt 3 \\
= \sqrt {{{\sqrt 3 }^2} - 2.\sqrt 3 .1 + {1^2}} - \sqrt 3 \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - \sqrt 3 \\
= \left| {\sqrt 3 - 1} \right| - \sqrt 3 \\
= \left( {\sqrt 3 - 1} \right) - \sqrt 3 \\
= - 1\\
d,\\
D = 3 - x + \sqrt {{x^2} + 6x + 9} \\
= 3 - x + \sqrt {{x^2} + 2.x.3 + {3^2}} \\
= 3 - x + \sqrt {{{\left( {x + 3} \right)}^2}} \\
= 3 - x + \left| {x + 3} \right|\\
TH1:\,\,\,x + 3 \ge 0 \Leftrightarrow x \ge - 3\\
\Rightarrow \left| {x + 3} \right| = x + 3\\
\Rightarrow D = 3 - x + x + 3 = 6\\
TH2:\,\,\,\,x + 3 < 0 \Leftrightarrow x < - 3\\
\Rightarrow \left| {x + 3} \right| = - \left( {x + 3} \right)\\
\Rightarrow D = 3 - x - \left( {x + 3} \right) = 3 - x - x - 3 = - 2x\\
\Rightarrow \left[ \begin{array}{l}
D = 6,\,\,\,\,x \ge - 3\\
D = - 2x,\,\,\,\,x < - 3
\end{array} \right.
\end{array}\)
