Đáp án:
\(\begin{array}{l}
1,\\
a,\,\,\,\,\dfrac{{63}}{{100}}\\
b,\,\,\,\,\dfrac{9}{5}\\
c,\,\,\,\,\dfrac{{13}}{2}\\
d,\,\,\,\,108\\
2,\\
A = \dfrac{{7\sqrt {10} }}{2}\\
3,\\
a,\,\,\,\,3x - 15\\
b,\,\,\,\,{x^2} - 2x\\
c,\,\,\,\,3x\\
d,\,\,\,\, - \dfrac{1}{{4x}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\sqrt {1\dfrac{{24}}{{25}}.5\dfrac{1}{{16}}.0,01} = \sqrt {\dfrac{{49}}{{25}}.\dfrac{{81}}{{16}}.\dfrac{1}{{100}}} = \sqrt {\dfrac{{{7^2}}}{{{5^2}}}.\dfrac{{{9^2}}}{{{4^2}}}.\dfrac{{{1^2}}}{{{{10}^2}}}} \\
= \sqrt {{{\left( {\dfrac{7}{5}.\dfrac{9}{4}.\dfrac{1}{{10}}} \right)}^2}} = \sqrt {{{\left( {\dfrac{{63}}{{100}}} \right)}^2}} = \dfrac{{63}}{{100}}\\
b,\\
\sqrt {2,25.1,46 - 2,25.0,02} = \sqrt {2,25.\left( {1,46 - 0,02} \right)} \\
= \sqrt {2,25.1,44} = \sqrt {\dfrac{{225}}{{100}}.\dfrac{{144}}{{100}}} = \sqrt {\dfrac{{{{15}^2}}}{{100}}.\dfrac{{{{12}^2}}}{{100}}} \\
= \sqrt {{{\left( {\dfrac{{15.12}}{{100}}} \right)}^2}} = \sqrt {{{\left( {\dfrac{9}{5}} \right)}^2}} = \dfrac{9}{5}\\
c,\\
\sqrt {2,5.16,9} = \sqrt {\dfrac{{25}}{{10}}.\dfrac{{169}}{{10}}} = \sqrt {\dfrac{{{5^2}{{.13}^2}}}{{{{10}^2}}}} = \sqrt {{{\left( {\dfrac{{5.13}}{{10}}} \right)}^2}} \\
= \sqrt {{{\left( {\dfrac{{13}}{2}} \right)}^2}} = \dfrac{{13}}{2}\\
d,\\
\sqrt {{{117,5}^2} - {{26,5}^2} - 1440} \\
= \sqrt {\left( {117,5 - 26,5} \right).\left( {117,5 + 26,5} \right) - 1440} \\
= \sqrt {91.144 - 1440} \\
= \sqrt {144.\left( {91 - 10} \right)} \\
= \sqrt {144.81} = \sqrt {{{12}^2}{{.9}^2}} \\
= \sqrt {{{\left( {12.9} \right)}^2}} = \sqrt {{{108}^2}} = 108\\
2,\\
A = \sqrt {0,1} + \sqrt {0,9} + \sqrt {6,4} + \sqrt {0,4} + \sqrt {44,1} \\
= \sqrt {\dfrac{1}{{10}}} + \sqrt {\dfrac{9}{{10}}} + \sqrt {\dfrac{{64}}{{10}}} + \sqrt {\dfrac{4}{{10}}} + \sqrt {\dfrac{{441}}{{10}}} \\
= \dfrac{1}{{\sqrt {10} }}\left( {\sqrt 1 + \sqrt 9 + \sqrt {64} + \sqrt 4 + \sqrt {441} } \right)\\
= \dfrac{1}{{\sqrt {10} }}.\left( {\sqrt {{1^2}} + \sqrt {{3^2}} + \sqrt {{8^2}} + \sqrt {{2^2}} + {{\sqrt {21} }^2}} \right)\\
= \dfrac{1}{{\sqrt {10} }}.\left( {1 + 3 + 8 + 2 + 21} \right)\\
= \dfrac{1}{{\sqrt {10} }}.35\\
= \dfrac{{35\sqrt {10} }}{{10}}\\
= \dfrac{{7\sqrt {10} }}{2}\\
3,\\
a,\\
x \ge 5 \Rightarrow x - 5 \ge 0 \Rightarrow \left| {x - 5} \right| = x - 5\\
\sqrt {9{{\left( {x - 5} \right)}^2}} = \sqrt 9 .\sqrt {{{\left( {x - 5} \right)}^2}} = \sqrt {{3^2}} .\left| {x - 5} \right| = 3.\left( {x - 5} \right) = 3x - 15\\
b,\\
x < 0 \Rightarrow \left\{ \begin{array}{l}
x < 0\\
x - 2 < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left| x \right| = - x\\
\left| {x - 2} \right| = - \left( {x - 2} \right)
\end{array} \right.\\
\sqrt {{x^2}{{\left( {x - 2} \right)}^2}} = \sqrt {{x^2}} .\sqrt {{{\left( {x - 2} \right)}^2}} = \left| x \right|.\left| {x - 2} \right|\\
= \left( { - x} \right).\left[ { - \left( {x - 2} \right)} \right] = x.\left( {x - 2} \right) = {x^2} - 2x\\
c,\\
x > 0 \Rightarrow \left| x \right| = x\\
\dfrac{{\sqrt {108{x^3}} }}{{\sqrt {12x} }} = \sqrt {\dfrac{{108{x^3}}}{{12x}}} = \sqrt {9{x^2}} = \sqrt 9 .\sqrt {{x^2}} = \sqrt {{3^2}} .\left| x \right| = 3x\\
d,\\
x < 0 \Rightarrow \left| x \right| = - x\\
\dfrac{{\sqrt {13{x^4}{y^6}} }}{{\sqrt {208{x^6}{y^6}} }} = \sqrt {\dfrac{{13{x^4}{y^6}}}{{208{x^6}{y^6}}}} = \sqrt {\dfrac{1}{{16{x^2}}}} = \sqrt {\dfrac{1}{{{4^2}.{x^2}}}} \\
= \dfrac{1}{{\left| {4x} \right|}} = \dfrac{1}{{4\left| x \right|}} = \dfrac{1}{{4.\left( { - x} \right)}} = - \dfrac{1}{{4x}}
\end{array}\)
