Đáp án:
\(\begin{array}{l}
1,\\
a,\,\,\,\sqrt 7 .\left( {\sqrt 7 + 1} \right)\\
b,\,\,\,\sqrt y .\left( {\sqrt y - 3} \right)\\
c,\,\,\,\sqrt 7 .\left( {\sqrt 3 + \sqrt 5 } \right)\\
d,\,\,\,\sqrt {21} .\left( {\sqrt 3 - \sqrt 7 } \right)\\
2,\\
a,\,\,\,\left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)\\
b,\,\,\,\left( {\sqrt 3 - \sqrt x } \right)\left( {\sqrt 3 + \sqrt x } \right)\\
c,\,\,\,\left( {4a - \sqrt 5 } \right)\left( {4a + \sqrt 5 } \right)\\
d,\,\,\,{\left( {\sqrt x - 3} \right)^2}\\
e,\,\,\,{\left( {\sqrt {x + 5} - 2} \right)^2}\\
3,\\
a,\,\,\,\dfrac{{ - 14}}{5}\\
b,\,\,\, - 11\sqrt 3 \\
c,\,\,\, - \dfrac{3}{2}\\
d,\,\,\,10\\
e,\,\,\,5\\
f,\,\,\,4\\
g,\,\,\, - 1\\
h,\,\,\,\,1
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
7 + \sqrt 7 = {\sqrt 7 ^2} + \sqrt 7 = \sqrt 7 .\left( {\sqrt 7 + 1} \right)\\
b,\\
y - 3\sqrt y = {\sqrt y ^2} - 3\sqrt y = \sqrt y .\left( {\sqrt y - 3} \right)\\
c,\\
\sqrt {21} + \sqrt {35} = \sqrt {7.3} + \sqrt {7.5} = \sqrt 7 .\sqrt 3 + \sqrt 7 .\sqrt 5 = \sqrt 7 .\left( {\sqrt 3 + \sqrt 5 } \right)\\
d,\\
3\sqrt 7 - 7\sqrt 3 = {\sqrt 3 ^2}.\sqrt 7 - {\sqrt 7 ^2}.\sqrt 3 \\
= \sqrt 3 .\left( {\sqrt 3 .\sqrt 7 } \right) - \sqrt 7 .\left( {\sqrt 7 .\sqrt 3 } \right)\\
= \sqrt 3 .\sqrt {3.7} - \sqrt 7 .\sqrt {7.3} \\
= \sqrt 3 .\sqrt {21} - \sqrt 7 .\sqrt {21} \\
= \sqrt {21} .\left( {\sqrt 3 - \sqrt 7 } \right)\\
2,\\
a,\\
a - 9 = {\sqrt a ^2} - {3^2} = \left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)\\
b,\\
3 - x = {\sqrt 3 ^2} - {\sqrt x ^2} = \left( {\sqrt 3 - \sqrt x } \right)\left( {\sqrt 3 + \sqrt x } \right)\\
c,\\
16{a^2} - 5 = {\left( {4a} \right)^2} - {\sqrt 5 ^2} = \left( {4a - \sqrt 5 } \right)\left( {4a + \sqrt 5 } \right)\\
d,\\
x - 6\sqrt x + 9 = {\sqrt x ^2} - 2.\sqrt x .3 + {3^2} = {\left( {\sqrt x - 3} \right)^2}\\
e,\\
x + 9 - 4\sqrt {x + 5} = \left( {x + 5} \right) - 4\sqrt {x + 5} + 4\\
= {\sqrt {x + 5} ^2} - 2.\sqrt {x + 5} .2 + {2^2}\\
= {\left( {\sqrt {x + 5} - 2} \right)^2}\\
3,\\
a,\\
\sqrt {0,04} + \sqrt {0,36} - 4\sqrt {0,81} \\
= \sqrt {\dfrac{4}{{100}}} + \sqrt {\dfrac{{36}}{{100}}} - 4\sqrt {\dfrac{{81}}{{100}}} \\
= \sqrt {\dfrac{{{2^2}}}{{{{10}^2}}}} + \sqrt {\dfrac{{{6^2}}}{{{{10}^2}}}} - 4\sqrt {\dfrac{{{9^2}}}{{{{10}^2}}}} \\
= \dfrac{2}{{10}} + \dfrac{6}{{10}} - 4.\dfrac{9}{{10}}\\
= \dfrac{{2 + 6 - 4.9}}{{10}} = \dfrac{{ - 28}}{{10}} = \dfrac{{ - 14}}{5}\\
b,\\
3\sqrt 3 + \sqrt {12} - \sqrt {48} - 2\sqrt {108} \\
= 3\sqrt 3 + \sqrt {4.3} - \sqrt {16.3} - 2.\sqrt {36.3} \\
= 3\sqrt 3 + \sqrt {{2^2}.3} - \sqrt {{4^2}.3} - 2\sqrt {{6^2}.3} \\
= 3\sqrt 3 + 2\sqrt 3 - 4\sqrt 3 - 2.6\sqrt 3 \\
= 3\sqrt 3 + 2\sqrt 3 - 4\sqrt 3 - 12\sqrt 3 \\
= - 11\sqrt 3 \\
c,\\
\sqrt 9 - 2\sqrt {\dfrac{{25}}{4}} + \sqrt {\dfrac{1}{4}} \\
= \sqrt {{3^2}} - 2\sqrt {\dfrac{{{5^2}}}{{{2^2}}}} + \sqrt {\dfrac{{{1^2}}}{{{2^2}}}} \\
= 3 - 2.\dfrac{5}{2} + \dfrac{1}{2} = 3 - 5 + \dfrac{1}{2} = - \dfrac{3}{2}\\
d,\\
\left( {\sqrt {20} - \sqrt {45} + 3\sqrt 5 } \right).\sqrt 5 \\
= \left( {\sqrt {4.5} - \sqrt {9.5} + 3\sqrt 5 } \right).\sqrt 5 \\
= \left( {\sqrt {{2^2}.5} - \sqrt {{3^2}.5} + 3\sqrt 5 } \right).\sqrt 5 \\
= \left( {2\sqrt 5 - 3\sqrt 5 + 3\sqrt 5 } \right).\sqrt 5 \\
= 2\sqrt 5 .\sqrt 5 = 2.{\sqrt 5 ^2} = 2.5 = 10\\
e,\\
{\left( {\sqrt 5 } \right)^2} - \sqrt {4.3} + \sqrt 3 + \dfrac{3}{{\sqrt 3 }}\\
= 5 - \sqrt {{2^2}.3} + \sqrt 3 + \dfrac{{{{\sqrt 3 }^2}}}{{\sqrt 3 }}\\
= 5 - 2\sqrt 3 + \sqrt 3 + \sqrt 3 \\
= 5\\
f,\\
\sqrt {6 - 2\sqrt 5 } .\left( {\sqrt 5 + 1} \right)\\
= \sqrt {5 - 2\sqrt 5 + 1} .\left( {\sqrt 5 + 1} \right)\\
= \sqrt {{{\sqrt 5 }^2} - 2.\sqrt 5 .1 + {1^2}} .\left( {\sqrt 5 + 1} \right)\\
= \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} .\left( {\sqrt 5 + 1} \right)\\
= \left| {\sqrt 5 - 1} \right|.\left( {\sqrt 5 + 1} \right)\\
= \left( {\sqrt 5 - 1} \right).\left( {\sqrt 5 + 1} \right)\\
= {\sqrt 5 ^2} - {1^2} = 5 - 1 = 4\\
g,\\
\sqrt {50} - 3\sqrt 8 + \sqrt {3 - 2\sqrt 2 } \\
= \sqrt {25.2} - 3\sqrt {4.2} + \sqrt {2 - 2\sqrt 2 + 1} \\
= \sqrt {{5^2}.2} - 3.\sqrt {{2^2}.2} + \sqrt {{{\sqrt 2 }^2} - 2.\sqrt 2 .1 + {1^2}} \\
= 5\sqrt 2 - 3.2\sqrt 2 + \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \\
= 5\sqrt 2 - 6\sqrt 2 + \left| {\sqrt 2 - 1} \right|\\
= - \sqrt 2 + \left( {\sqrt 2 - 1} \right)\\
= - 1\\
h,\\
\left( {3\sqrt {32} - 2\sqrt {18} - \sqrt {50} } \right):\sqrt 2 \\
= \left( {3\sqrt {16.2} - 2\sqrt {9.2} - \sqrt {25.2} } \right):\sqrt 2 \\
= \left( {3\sqrt {{4^2}.2} - 2\sqrt {{3^2}.2} - \sqrt {{5^2}.2} } \right):\sqrt 2 \\
= \left( {3.4\sqrt 2 - 2.3\sqrt 2 - 5\sqrt 2 } \right):\sqrt 2 \\
= \left( {12\sqrt 2 - 6\sqrt 2 - 5\sqrt 2 } \right):\sqrt 2 \\
= \sqrt 2 :\sqrt 2 \\
= 1
\end{array}\)
