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Mk năm nay mới lên lớp 9 thôi nhưng cũng biết chút!Mk giải ho bạn câu 1 còn lại bạn tự giải nhé!
1,\(\frac{1}{1+\sqrt{5}}\)+\(\frac{1}{\sqrt{5}-1}\)
=\(\frac{1}{\sqrt{5}+1}\)+\(\frac{1}{\sqrt{5}-1}\)
=\(\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)+\(\frac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
=\(\frac{\sqrt{5}-1}{5-1}\)+\(\frac{\sqrt{5}+1}{5-1}\)
=\(\frac{\sqrt{5}-1}{4}\)+\(\frac{\sqrt{5}+1}{4}\)
=\(\frac{\sqrt{5}-1+\sqrt{5}+1}{4}\)
=\(\frac{2\sqrt{5}}{4}\)
=\(\frac{\sqrt{5}}{2}\)
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
Em thử nhá, ko chắc đâu
1) \(\frac{2}{\sqrt{20}}=\frac{2\sqrt{20}}{20}\) 2) \(\frac{4}{\sqrt{8}}=\frac{4\sqrt{8}}{8}\)
3) \(\frac{2+\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{2}+\sqrt{6}}{2}\) 4) \(\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}+2}{6-4}=\frac{\sqrt{6}+2}{2}\)
5) \(\frac{1}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=-\left(\sqrt{2}+\sqrt{3}\right)\)
6) \(\frac{9a-b}{3\sqrt{a}-\sqrt{b}}=\frac{\left(9a-b\right)\left(3\sqrt{a}+b\right)}{\left(3\sqrt{a}-\sqrt{b}\right)\left(3\sqrt{a}+\sqrt{b}\right)}=\left(3\sqrt{a}+b\right)\)
7) + 8) em chưa nghĩ ra
ong tth :v
\(\frac{2}{\sqrt{20}}=\frac{\sqrt{4}}{\sqrt{4}.\sqrt{5}}=\frac{1}{\sqrt{5}}\)
\(\frac{4}{\sqrt{8}}=\frac{\sqrt{16}}{\sqrt{8}}=\sqrt{2}\)
\(\frac{2+\sqrt{3}}{\sqrt{2}}=\sqrt{2}+\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{2}+\sqrt{1,5}\)
\(\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}+2}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}=\frac{\sqrt{6}+2}{2}\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{3}+\sqrt{2}}{-1}=-\sqrt{3}-\sqrt{2}\)
7: chưa
8: chưa
9:\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(2+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
\(1)\dfrac{{14}}{{\sqrt 7 }} = \dfrac{{14\sqrt 7 }}{{\sqrt 7 .\sqrt 7 }} = \dfrac{{14\sqrt 7 }}{7} = 2\sqrt 7 \\ 2)\dfrac{{\sqrt 3 }}{{\sqrt 2 }} = \dfrac{{\sqrt 3 .\sqrt 2 }}{{\sqrt 2 .\sqrt 2 }} = \dfrac{{\sqrt 6 }}{2}\\ 3)\dfrac{5}{{\sqrt {10} }} = \dfrac{{5\sqrt {10} }}{{\sqrt {10} .\sqrt {10} }} = \dfrac{{5\sqrt {10} }}{{10}} = \dfrac{{\sqrt {10} }}{2}\\ 4)\dfrac{3}{{2\sqrt 5 }} = \dfrac{{3.2\sqrt 5 }}{{2\sqrt 5 .2\sqrt 5 }} = \dfrac{{6\sqrt 5 }}{{20}} = \dfrac{{3\sqrt 5 }}{{10}}\\ 5)\dfrac{{7 + \sqrt 7 }}{{\sqrt 7 + 1}} = \dfrac{{\left( {7 + \sqrt 7 } \right)\left( {\sqrt 7 - 1} \right)}}{{\left( {\sqrt 7 + 1} \right)\left( {\sqrt 7 - 1} \right)}} = \dfrac{{6\sqrt 7 }}{6} = \sqrt 7 \\ 6)\dfrac{{\sqrt 2 - \sqrt 6 }}{{3\sqrt 3 - 3}} = \dfrac{{\left( {\sqrt 2 - \sqrt 6 } \right)\left( {3\sqrt 3 + 3} \right)}}{{\left( {3\sqrt 3 - 3} \right)\left( {3\sqrt 3 + 3} \right)}} = \dfrac{{ - 2\sqrt 2 }}{6} = \dfrac{{ - \sqrt 2 }}{3}\\ 7)\dfrac{{\sqrt 3 }}{{3 - \sqrt 3 }} = \dfrac{{\sqrt 3 \left( {3 + \sqrt 3 } \right)}}{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}} = \dfrac{{3\sqrt 3 + 3}}{6} = \dfrac{{3\left( {\sqrt 3 + 1} \right)}}{6} = \dfrac{{\sqrt 3 + 1}}{2}\\ 8)\dfrac{2}{{2 - \sqrt 3 }} = \dfrac{{2\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 4 + 2\sqrt 3 \\ 9)\dfrac{{\sqrt 3 + 2}}{{2 - \sqrt 3 }} = \dfrac{{\left( {\sqrt 3 + 2} \right)\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 7 + 4\sqrt 3 \\ 10)\dfrac{{3\sqrt 5 }}{{2\sqrt 5 - 1}} = \dfrac{{3\sqrt 5 \left( {2\sqrt 5 + 1} \right)}}{{\left( {2\sqrt 5 - 1} \right)\left( {2\sqrt 5 + 1} \right)}} = \dfrac{{30 + 3\sqrt 5 }}{{19}}\\ 11)\dfrac{1}{{\sqrt 3 }} = \dfrac{{1.\sqrt 3 }}{{\sqrt 3 .\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3} \)