Mọi người ơi giúp em bài này với
Ai đã từng học lớp 9 rồi thì giúp em pls
Trục căn thức
1, \(\frac{14}{\sqrt{7}}\)
2, \(\frac{\sqrt{3}}{\sqrt{2}}\)
3,\(\frac{5}{\sqrt{10}}\)
4, \(\frac{3}{2\sqrt{5}}\)
5, \(\frac{7+\sqrt{7}}{\sqrt{7}+1}\)
6, \(\frac{\sqrt{2}-\sqrt{6}}{3\sqrt{3}-3}\)
7, \(\frac{\sqrt{3}}{3-\sqrt{3}}\)
8, \(\frac{2}{2-\sqrt{3}}\)
9, \(\frac{\sqrt{3}+2}{2-\sqrt{3}}\)
10, \(\frac{3\sqrt{5}}{2\sqrt{5}-1}\)
11, \(\frac{1}{\sqrt{3}}\)
\(1)\dfrac{{14}}{{\sqrt 7 }} = \dfrac{{14\sqrt 7 }}{{\sqrt 7 .\sqrt 7 }} = \dfrac{{14\sqrt 7 }}{7} = 2\sqrt 7 \\ 2)\dfrac{{\sqrt 3 }}{{\sqrt 2 }} = \dfrac{{\sqrt 3 .\sqrt 2 }}{{\sqrt 2 .\sqrt 2 }} = \dfrac{{\sqrt 6 }}{2}\\ 3)\dfrac{5}{{\sqrt {10} }} = \dfrac{{5\sqrt {10} }}{{\sqrt {10} .\sqrt {10} }} = \dfrac{{5\sqrt {10} }}{{10}} = \dfrac{{\sqrt {10} }}{2}\\ 4)\dfrac{3}{{2\sqrt 5 }} = \dfrac{{3.2\sqrt 5 }}{{2\sqrt 5 .2\sqrt 5 }} = \dfrac{{6\sqrt 5 }}{{20}} = \dfrac{{3\sqrt 5 }}{{10}}\\ 5)\dfrac{{7 + \sqrt 7 }}{{\sqrt 7 + 1}} = \dfrac{{\left( {7 + \sqrt 7 } \right)\left( {\sqrt 7 - 1} \right)}}{{\left( {\sqrt 7 + 1} \right)\left( {\sqrt 7 - 1} \right)}} = \dfrac{{6\sqrt 7 }}{6} = \sqrt 7 \\ 6)\dfrac{{\sqrt 2 - \sqrt 6 }}{{3\sqrt 3 - 3}} = \dfrac{{\left( {\sqrt 2 - \sqrt 6 } \right)\left( {3\sqrt 3 + 3} \right)}}{{\left( {3\sqrt 3 - 3} \right)\left( {3\sqrt 3 + 3} \right)}} = \dfrac{{ - 2\sqrt 2 }}{6} = \dfrac{{ - \sqrt 2 }}{3}\\ 7)\dfrac{{\sqrt 3 }}{{3 - \sqrt 3 }} = \dfrac{{\sqrt 3 \left( {3 + \sqrt 3 } \right)}}{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}} = \dfrac{{3\sqrt 3 + 3}}{6} = \dfrac{{3\left( {\sqrt 3 + 1} \right)}}{6} = \dfrac{{\sqrt 3 + 1}}{2}\\ 8)\dfrac{2}{{2 - \sqrt 3 }} = \dfrac{{2\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 4 + 2\sqrt 3 \\ 9)\dfrac{{\sqrt 3 + 2}}{{2 - \sqrt 3 }} = \dfrac{{\left( {\sqrt 3 + 2} \right)\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 7 + 4\sqrt 3 \\ 10)\dfrac{{3\sqrt 5 }}{{2\sqrt 5 - 1}} = \dfrac{{3\sqrt 5 \left( {2\sqrt 5 + 1} \right)}}{{\left( {2\sqrt 5 - 1} \right)\left( {2\sqrt 5 + 1} \right)}} = \dfrac{{30 + 3\sqrt 5 }}{{19}}\\ 11)\dfrac{1}{{\sqrt 3 }} = \dfrac{{1.\sqrt 3 }}{{\sqrt 3 .\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3} \)
