Đáp án:
\(\begin{array}{l}
1,\\
a,\,\,\,\,\,4\\
b,\,\,\,\,\,20\\
c,\,\,\,\,\,6\\
d,\,\,\,\,\,3\\
2,\\
a,\,\,\,\,\,21\\
b,\,\,\,\,\,24\\
c,\,\,\,\,\,13\\
d,\,\,\,\,\dfrac{{3\sqrt {10} }}{5}\\
3,\\
a,\,\,\,\,6\\
b,\,\,\,\,12\\
4,\\
a,\\
x = 14\\
b,\\
x = 3\\
5,\\
a,\,\,\,\,\,2a - 6\\
b,\,\,\,\,\,{b^2} + b\\
6,\\
a,\,\,\,\,\,3\\
b,\,\,\,\,\,12\\
c,\,\,\,\,\sqrt {10} \\
7,\\
\,\sqrt {7 + 2\sqrt {10} } - \sqrt 5 = \sqrt 2 \\
8,\\
a,\,\,\,\,x - \sqrt 5 \\
b,\,\,\,\,\dfrac{{x + \sqrt 2 }}{{x - \sqrt 2 }}\\
c,\,\,\,\,\dfrac{{\sqrt 2 }}{2}\\
d,\,\,\,\,1 + \sqrt 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\sqrt 2 .\sqrt 8 = \sqrt {2.8} = \sqrt {16} = \sqrt {{4^2}} = \left| 4 \right| = 4\\
b,\\
\sqrt {10} .\sqrt {40} = \sqrt {10.40} = \sqrt {400} = \sqrt {{{20}^2}} = \left| {20} \right| = 20\\
c,\\
\sqrt 2 .\sqrt 3 .\sqrt 6 = \sqrt {2.3} .\sqrt 6 = \sqrt 6 .\sqrt 6 = {\sqrt 6 ^2} = 6\\
d,\\
\sqrt {10} .\sqrt {0,9} = \sqrt {10.0,9} = \sqrt 9 = \sqrt {{3^2}} = \left| 3 \right| = 3\\
2,\\
a,\\
\sqrt {9.49} = \sqrt {{3^2}{{.7}^2}} = \sqrt {{{\left( {3.7} \right)}^2}} = \sqrt {{{21}^2}} = \left| {21} \right| = 21\\
b,\\
\sqrt {90.6,4} = \sqrt {9.10.6,4} = \sqrt {9.64} = \sqrt {{3^2}{{.8}^2}} \\
= \sqrt {{{\left( {3.8} \right)}^2}} = \sqrt {{{24}^2}} = \left| {24} \right| = 24\\
c,\\
\sqrt {4.1,69,25} = \sqrt {{2^2}{{.1,3}^2}{{.5}^2}} = \sqrt {{{\left( {2.1,3.5} \right)}^2}} \\
= \sqrt {{{13}^2}} = \left| {13} \right| = 13\\
d,\\
\sqrt {2,5.1,44} = \sqrt {\dfrac{{25}}{{10}}.\dfrac{{144}}{{100}}} = \sqrt {\dfrac{{{5^2}}}{{10}}.\dfrac{{{{12}^2}}}{{{{10}^2}}}} \\
= \sqrt {{{\left( {\dfrac{{5.12}}{{10}}} \right)}^2}.\dfrac{1}{{10}}} = \sqrt {{6^2}.\dfrac{1}{{10}}} = 6.\dfrac{1}{{\sqrt {10} }} = \dfrac{{6\sqrt {10} }}{{10}} = \dfrac{{3\sqrt {10} }}{5}\\
3,\\
a,\\
\sqrt {{{6,8}^2} - {{3,2}^2}} = \sqrt {\left( {6,8 - 3,2} \right)\left( {6,8 + 3,2} \right)} \\
= \sqrt {3,6.10} = \sqrt {36} = \sqrt {{6^2}} = \left| 6 \right| = 6\\
b,\\
\sqrt {{{21,8}^2} - {{18,2}^2}} = \sqrt {\left( {21,8 + 18,2} \right).\left( {21,8 - 18,2} \right)} \\
= \sqrt {40.3,6} = \sqrt {4.10.3,6} = \sqrt {4.36} = \sqrt {{2^2}{{.6}^2}} \\
= \sqrt {{{\left( {2.6} \right)}^2}} = \sqrt {{{12}^2}} = \left| {12} \right| = 12\\
4,\\
a,\\
DKXD:\,\,\,x - 5 \ge 0 \Leftrightarrow x \ge 5\\
\sqrt {x - 5} = 3\\
\Leftrightarrow x - 5 = {3^2}\\
\Leftrightarrow x - 5 = 9\\
\Leftrightarrow x = 9 + 5\\
\Leftrightarrow x = 14\\
b,\\
DKXD:\,\,\,2x - 1 \ge 0 \Leftrightarrow x \ge \dfrac{1}{2}\\
\sqrt {2x - 1} = \sqrt 5 \\
\Leftrightarrow 2x - 1 = {\sqrt 5 ^2}\\
\Leftrightarrow 2x - 1 = 5\\
\Leftrightarrow 2x = 6\\
\Leftrightarrow x = 3\\
5,\\
a,\\
a \ge 3 \Rightarrow a - 3 \ge 0 \Rightarrow \left| {a - 3} \right| = a - 3\\
\sqrt {4{{\left( {a - 3} \right)}^2}} = \sqrt {{2^2}.{{\left( {a - 3} \right)}^2}} = \sqrt {{2^2}} .\sqrt {{{\left( {a - 3} \right)}^2}} \\
= \left| 2 \right|.\left| {a - 3} \right| = 2.\left( {a - 3} \right) = 2a - 6\\
b,\\
b > 0 \Rightarrow \left\{ \begin{array}{l}
b > 0\\
b + 1 > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| b \right| = b\\
\left| {b + 1} \right| = b + 1
\end{array} \right.\\
\sqrt {{b^2}.{{\left( {b + 1} \right)}^2}} = \sqrt {{b^2}} .\sqrt {{{\left( {b + 1} \right)}^2}} \\
= \left| b \right|.\left| {b + 1} \right| = b.\left( {b + 1} \right) = {b^2} + b\\
6,\\
a,\\
\sqrt {4 + \sqrt 7 } .\sqrt {4 - \sqrt 7 } \\
= \sqrt {\left( {4 + \sqrt 7 } \right).\left( {4 - \sqrt 7 } \right)} \\
= \sqrt {{4^2} - {{\sqrt 7 }^2}} \\
= \sqrt {16 - 7} \\
= \sqrt 9 \\
= \sqrt {{3^2}} \\
= 3\\
b,\\
2\sqrt 2 .\left( {\sqrt 3 - \sqrt {18} } \right) - 2\sqrt 6 \\
= 2\sqrt 2 .\sqrt 3 - 2\sqrt 2 .\sqrt {18} - 2\sqrt 6 \\
= 2\sqrt {2.3} - 2.\sqrt {2.18} - 2\sqrt 6 \\
= 2\sqrt 6 - 2.\sqrt {36} - 2\sqrt 6 \\
= 2\sqrt {36} \\
= 2\sqrt {{6^2}} \\
= 2.6\\
= 12\\
c,\\
\sqrt {3 + \sqrt 5 } + \sqrt {3 - \sqrt 5 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 2 .\sqrt {3 + \sqrt 5 } + \sqrt 2 .\sqrt {3 - \sqrt 5 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {6 + 2\sqrt 5 } + \sqrt {6 - 2\sqrt 5 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {5 + 2.\sqrt 5 .1 + 1} + \sqrt {5 - 2.\sqrt 5 .1 + 1} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\left| {\sqrt 5 + 1} \right| + \left| {\sqrt 5 - 1} \right|} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 5 + 1 + \sqrt 5 - 1} \right)\\
= \dfrac{1}{{\sqrt 2 }}.2\sqrt 5 \\
= \sqrt 2 .\sqrt 5 \\
= \sqrt {10} \\
7,\\
\sqrt {7 + 2\sqrt {10} } - \sqrt 5 \\
= \sqrt {5 + 2\sqrt {10} + 2} - \sqrt 5 \\
= \sqrt {{{\sqrt 5 }^2} + 2.\sqrt 5 .\sqrt 2 + {{\sqrt 2 }^2}} - \sqrt 5 \\
= \sqrt {{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}} - \sqrt 5 \\
= \left| {\sqrt 5 + \sqrt 2 } \right| - \sqrt 5 \\
= \left( {\sqrt 5 + \sqrt 2 } \right) - \sqrt 5 \\
= \sqrt 5 + \sqrt 2 - \sqrt 5 \\
= \sqrt 2 \\
8,\\
a,\\
\dfrac{{{x^2} - 5}}{{x + \sqrt 5 }} = \dfrac{{{x^2} - {{\sqrt 5 }^2}}}{{x + \sqrt 5 }} = \dfrac{{\left( {x - \sqrt 5 } \right)\left( {x + \sqrt 5 } \right)}}{{x + \sqrt 5 }} = x - \sqrt 5 \\
b,\\
\dfrac{{{x^2} + 2\sqrt 2 x + 2}}{{{x^2} - 2}} = \dfrac{{{x^2} + 2.x.\sqrt 2 + {{\sqrt 2 }^2}}}{{{x^2} - {{\sqrt 2 }^2}}}\\
= \dfrac{{{{\left( {x + \sqrt 2 } \right)}^2}}}{{\left( {x - \sqrt 2 } \right)\left( {x + \sqrt 2 } \right)}} = \dfrac{{x + \sqrt 2 }}{{x - \sqrt 2 }}\\
c,\\
\dfrac{{\sqrt 6 + \sqrt {14} }}{{2\sqrt 3 + \sqrt {28} }} = \dfrac{{\sqrt {2.3} + \sqrt {2.7} }}{{2\sqrt 3 + \sqrt {4.7} }} = \dfrac{{\sqrt 2 .\sqrt 3 + \sqrt 2 .\sqrt 7 }}{{2\sqrt 3 + \sqrt 4 .\sqrt 7 }}\\
= \dfrac{{\sqrt 2 .\sqrt 3 + \sqrt 2 .\sqrt 7 }}{{2\sqrt 3 + 2.\sqrt 7 }} = \dfrac{{\sqrt 2 .\left( {\sqrt 3 + \sqrt 7 } \right)}}{{2.\left( {\sqrt 3 + \sqrt 7 } \right)}} = \dfrac{{\sqrt 2 }}{2}\\
d,\\
\dfrac{{\sqrt 2 + \sqrt 3 + \sqrt 6 + \sqrt 8 + \sqrt {16} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}\\
= \dfrac{{\sqrt 2 + \sqrt 3 + \sqrt 6 + \sqrt {{2^2}.2} + \sqrt {{4^2}} }}{{\sqrt 2 + \sqrt 3 + \sqrt {{2^2}} }}\\
= \dfrac{{\sqrt 2 + \sqrt 3 + \sqrt 6 + 2\sqrt 2 + 4}}{{\sqrt 2 + \sqrt 3 + 2}}\\
= \dfrac{{\left( {\sqrt 2 + \sqrt 3 + 2} \right) + \left( {\sqrt 6 + 2\sqrt 2 + 2} \right)}}{{\sqrt 2 + \sqrt 3 + 2}}\\
= \dfrac{{\left( {2 + \sqrt 3 + 2} \right) + \sqrt 2 .\left( {\sqrt 3 + 2 + \sqrt 2 } \right)}}{{\sqrt 2 + \sqrt 3 + 2}}\\
= \dfrac{{\left( {\sqrt 2 + \sqrt 3 + 2} \right).\left( {1 + \sqrt 2 } \right)}}{{\sqrt 2 + \sqrt 3 + 2}}\\
= 1 + \sqrt 2
\end{array}\)
