Đáp án:
\(\begin{array}{l}
20,\\
a,\,\,\,5{a^2}\\
b,\,\,\,x\\
c,\,\,\, - 1\\
d,\,\,\,\,6\\
21,\\
a,\,\,\,25\\
b,\,\,\,\,\dfrac{{\sqrt 5 }}{5}\\
c,\,\,\,\,2 - 2\sqrt 6 \\
d,\,\,\, - 2\\
22,\\
a,\,\,\,\sqrt {10} \\
b,\,\,\,\dfrac{{13\sqrt 3 }}{{15}}\\
c,\,\,\,4\sqrt 7 - \sqrt 2 \\
d,\,\,\, - 4\sqrt 3
\end{array}\)
\(\begin{array}{l}
23,\\
a,\,\,\,\dfrac{1}{2}\\
b,\,\,\,2\\
c,\,\,\,\,0\\
d,\,\,\, - 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
20,\\
a,\\
\sqrt {9{a^4}} + 2{a^2} = \sqrt {{3^2}.{{\left( {{a^2}} \right)}^2}} + 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}\\
b,\\
x \ge 0 \Rightarrow \left| x \right| = x\\
\sqrt {9{x^2}} - 2x = \sqrt {{3^2}.{x^2}} - 2x = \sqrt {{{\left( {3x} \right)}^2}} - 2x\\
= \left| {3x} \right| - 2x = 3x - 2x = x\\
c,\\
\sqrt {4 - 2\sqrt 3 } - \sqrt 3 = \sqrt {3 - 2\sqrt 3 + 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,\\
x > - 3 \Leftrightarrow x + 3 > 0 \Rightarrow \left| {x + 3} \right| = x + 3\\
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|\\
= 3 - x + \left( {x + 3} \right) = 6\\
21,\\
a,\\
\left( {2\sqrt {45} + \sqrt {80} - \sqrt {125} } \right).\sqrt 5 \\
= \left( {2\sqrt {9.5} + \sqrt {16.5} - \sqrt {25.5} } \right).\sqrt 5 \\
= \left( {2.\sqrt {{3^2}.5} + \sqrt {{4^2}.5} - \sqrt {{5^2}.5} } \right).\sqrt 5 \\
= \left( {2.3\sqrt 5 + 4\sqrt 5 - 5\sqrt 5 } \right).\sqrt 5 \\
= \left( {6\sqrt 5 + 4\sqrt 5 - 5\sqrt 5 } \right).\sqrt 5 \\
= 5\sqrt 5 .\sqrt 5 \\
= 5.5 = 25\\
b,\\
2\sqrt {\dfrac{{16}}{5}} - 3\sqrt {\dfrac{1}{{45}}} - 6\sqrt {\dfrac{4}{{20}}} \\
= 2.\sqrt {16.\dfrac{1}{5}} - 3.\sqrt {\dfrac{1}{9}.\dfrac{1}{5}} - 6\sqrt {\dfrac{1}{5}} \\
= 2.\sqrt {{4^2}.\dfrac{1}{5}} - 3.\sqrt {{{\left( {\dfrac{1}{3}} \right)}^2}.\dfrac{1}{5}} - 6\sqrt {\dfrac{1}{5}} \\
= 2.4.\sqrt {\dfrac{1}{5}} - 3.\dfrac{1}{3}.\sqrt {\dfrac{1}{5}} - 6\sqrt {\dfrac{1}{5}} \\
= 8\sqrt {\dfrac{1}{5}} - \sqrt {\dfrac{1}{5}} - 6\sqrt {\dfrac{1}{5}} \\
= \sqrt {\dfrac{1}{5}} = \dfrac{1}{{\sqrt 5 }} = \dfrac{{\sqrt 5 }}{5}\\
c,\\
3 + \sqrt {7 - 2\sqrt 6 } - 3\sqrt 6 \\
= 3 + \sqrt {6 - 2\sqrt 6 + 1} - 3\sqrt 6 \\
= 3 + \sqrt {{{\sqrt 6 }^2} - 2.\sqrt 6 .1 + {1^2}} - 3\sqrt 6 \\
= 3 + \sqrt {{{\left( {\sqrt 6 - 1} \right)}^2}} - 3\sqrt 6 \\
= 3 + \left| {\sqrt 6 - 1} \right| - 3\sqrt 6 \\
= 3 + \sqrt 6 - 1 - 3\sqrt 6 \\
= 2 - 2\sqrt 6 \\
d,\\
\dfrac{3}{{\sqrt 5 + \sqrt 2 }} - \dfrac{4}{{3 - \sqrt 5 }} + \dfrac{1}{{\sqrt 2 - 1}}\\
= \dfrac{{3.\left( {\sqrt 5 - \sqrt 2 } \right)}}{{\left( {\sqrt 5 + \sqrt 2 } \right).\left( {\sqrt 5 - \sqrt 2 } \right)}} - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}} + \dfrac{{\sqrt 2 + 1}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)}}\\
= \dfrac{{3.\left( {\sqrt 5 - \sqrt 2 } \right)}}{{{{\sqrt 5 }^2} - {{\sqrt 2 }^2}}} - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{{{3^2} - {{\sqrt 5 }^2}}} + \dfrac{{\sqrt 2 + 1}}{{{{\sqrt 2 }^2} - {1^2}}}\\
= \dfrac{{3.\left( {\sqrt 5 - \sqrt 2 } \right)}}{3} - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{4} + \dfrac{{\sqrt 2 + 1}}{1}\\
= \left( {\sqrt 5 - \sqrt 2 } \right) - \left( {3 + \sqrt 5 } \right) + \left( {\sqrt 2 + 1} \right)\\
= \sqrt 5 - \sqrt 2 - 3 - \sqrt 5 + \sqrt 2 + 1\\
= - 2\\
22,\\
a,\\
\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} \\
b,\\
2\sqrt {\dfrac{{16}}{3}} - 3\sqrt {\dfrac{1}{{27}}} - 6\sqrt {\dfrac{4}{{75}}} \\
= 2\sqrt {16.\dfrac{1}{3}} - 3\sqrt {\dfrac{1}{9}.\dfrac{1}{3}} - 6\sqrt {\dfrac{4}{{25}}.\dfrac{1}{3}} \\
= 2\sqrt {{4^2}.\dfrac{1}{3}} - 3\sqrt {{{\left( {\dfrac{1}{3}} \right)}^2}.\dfrac{1}{3}} - 6\sqrt {{{\left( {\dfrac{2}{5}} \right)}^2}.\dfrac{1}{3}} \\
= 2.4.\sqrt {\dfrac{1}{3}} - 3.\dfrac{1}{3}.\sqrt {\dfrac{1}{3}} - 6.\dfrac{2}{5}\sqrt {\dfrac{1}{3}} \\
= 8\sqrt {\dfrac{1}{3}} - \sqrt {\dfrac{1}{3}} - \dfrac{{12}}{5}\sqrt {\dfrac{1}{3}} \\
= \dfrac{{13}}{5}.\sqrt {\dfrac{1}{3}} \\
= \dfrac{{13}}{{5.\sqrt 3 }}\\
= \dfrac{{13\sqrt 3 }}{{15}}\\
c,\\
\dfrac{1}{{\sqrt 8 + \sqrt 7 }} + \sqrt {175} - \dfrac{{6\sqrt 2 - 4}}{{3 - \sqrt 2 }}\\
= \dfrac{{\sqrt 8 - \sqrt 7 }}{{\left( {\sqrt 8 + \sqrt 7 } \right)\left( {\sqrt 8 - \sqrt 7 } \right)}} + \sqrt {25.7} - \dfrac{{3.2\sqrt 2 - 2\sqrt 2 .\sqrt 2 }}{{3 - \sqrt 2 }}\\
= \dfrac{{\sqrt 8 - \sqrt 7 }}{{{{\sqrt 8 }^2} - {{\sqrt 7 }^2}}} + \sqrt {{5^2}.7} - \dfrac{{3\sqrt 2 \left( {3 - \sqrt 2 } \right)}}{{3 - \sqrt 2 }}\\
= \dfrac{{\sqrt 8 - \sqrt 7 }}{1} + 5\sqrt 7 - 3\sqrt 2 \\
= \sqrt 8 - \sqrt 7 + 5\sqrt 7 - 3\sqrt 2 \\
= 2\sqrt 2 + 4\sqrt 7 - 3\sqrt 2 \\
= 4\sqrt 7 - \sqrt 2 \\
d,\\
\sqrt {10 - \sqrt {84} } - \sqrt {34 + 2\sqrt {189} } \\
= \sqrt {10 - \sqrt {4.21} } - \sqrt {34 + 2\sqrt {189} } \\
= \sqrt {10 - 2.\sqrt {21} } - \sqrt {34 + 2\sqrt {189} } \\
= \sqrt {7 - 2.\sqrt 7 .\sqrt 3 + 3} - \sqrt {27 + 2.\sqrt {27} .\sqrt 7 + 7} \\
= \sqrt {{{\left( {\sqrt 7 - \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt {27} + \sqrt 7 } \right)}^2}} \\
= \left| {\sqrt 7 - \sqrt 3 } \right| - \left| {\sqrt {27} + \sqrt 7 } \right|\\
= \sqrt 7 - \sqrt 3 - \sqrt {27} - \sqrt 7 \\
= - \sqrt 3 - \sqrt {27} \\
= - \sqrt 3 - 3\sqrt 3 \\
= - 4\sqrt 3
\end{array}\)
\(\begin{array}{l}
23,\\
a,\\
\left( {\dfrac{2}{{\sqrt 3 - 1}} + \dfrac{3}{{\sqrt 3 - 2}} + \dfrac{{15}}{{3 - \sqrt 3 }}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\dfrac{{2.\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} + \dfrac{{3\left( {\sqrt 3 + 2} \right)}}{{\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 + 2} \right)}} + \dfrac{{15\left( {3 + \sqrt 3 } \right)}}{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\dfrac{{2.\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} + \dfrac{{3.\left( {\sqrt 3 + 2} \right)}}{{3 - 4}} + \dfrac{{15\left( {3 + \sqrt 3 } \right)}}{{9 - 3}}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\dfrac{{2.\left( {\sqrt 3 + 1} \right)}}{2} + \dfrac{{3.\left( {\sqrt 3 + 2} \right)}}{{ - 1}} + \dfrac{{15\left( {3 + \sqrt 3 } \right)}}{6}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left[ {\left( {\sqrt 3 + 1} \right) - 3\left( {\sqrt 3 + 2} \right) + \dfrac{5}{2}.\left( {3 + \sqrt 3 } \right)} \right].\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\sqrt 3 + 1 - 3\sqrt 3 - 6 + \dfrac{{15}}{2} + \dfrac{{5\sqrt 3 }}{2}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \dfrac{{5 + \sqrt 3 }}{2}.\dfrac{1}{{\sqrt 3 + 5}}\\
= \dfrac{1}{2}\\
b,\\
\dfrac{{\sqrt 3 }}{{\sqrt {\sqrt 3 + 1} - 1}} - \dfrac{{\sqrt 3 }}{{\sqrt {\sqrt 3 + 1} + 1}}\\
= \dfrac{{\sqrt 3 .\left( {\sqrt {\sqrt 3 + 1} + 1} \right) - \sqrt 3 .\left( {\sqrt {\sqrt 3 + 1} - 1} \right)}}{{\left( {\sqrt {\sqrt 3 + 1} - 1} \right).\left( {\sqrt {\sqrt 3 + 1} + 1} \right)}}\\
= \dfrac{{\sqrt 3 .\sqrt {\sqrt 3 + 1} + \sqrt 3 - \sqrt 3 .\sqrt {\sqrt 3 + 1} + \sqrt 3 }}{{{{\sqrt {\sqrt 3 + 1} }^2} - {1^2}}}\\
= \dfrac{{2\sqrt 3 }}{{\left( {\sqrt 3 + 1} \right) - 1}}\\
= \dfrac{{2\sqrt 3 }}{{\sqrt 3 }}\\
= 2\\
c,\\
\sqrt {\dfrac{3}{{20}}} + \sqrt {\dfrac{1}{{60}}} - 2\sqrt {\dfrac{1}{{15}}} \\
= \sqrt {\dfrac{9}{4}.\dfrac{1}{{15}}} + \sqrt {\dfrac{1}{4}.\dfrac{1}{{15}}} - 2\sqrt {\dfrac{1}{{15}}} \\
= \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2}.\dfrac{1}{{15}}} + \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2}.\dfrac{1}{{15}}} - 2\sqrt {\dfrac{1}{{15}}} \\
= \dfrac{3}{2}\sqrt {\dfrac{1}{{15}}} + \dfrac{1}{2}\sqrt {\dfrac{1}{{15}}} - 2\sqrt {\dfrac{1}{{15}}} \\
= 0\\
d,\\
\left( {\dfrac{{\sqrt {14} - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7 - \sqrt 5 }}\\
= \left( {\dfrac{{\sqrt 7 .\left( {\sqrt 2 - 1} \right)}}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 .\left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }}} \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \left( { - \sqrt 7 - \sqrt 5 } \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {\sqrt 7 + \sqrt 5 } \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {{{\sqrt 7 }^2} - {{\sqrt 5 }^2}} \right)\\
= - \left( {7 - 5} \right)\\
= - 2
\end{array}\)
