1. Trục căn thức ở mẫu:

\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)

=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)

\(=\frac{\sqrt{2009}-1}{4}\)

2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)

\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)

\(=6+3x\)

=> \(x^3-3x=6\)

=> \(B=x^3-3x+2000=6+2000=2006\)

\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)

\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)

\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)

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,\(\frac{1}{1+\sqrt{5}}\)+\(\frac{1}{\sqrt{5}-1}\)
=\(\frac{1}{\sqrt{5}+1}\)+\(\frac{1}{\sqrt{5}-1}\)
=\(\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)+\(\frac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
=\(\frac{\sqrt{5}-1}{5-1}\)+\(\frac{\sqrt{5}+1}{5-1}\)
=\(\frac{\sqrt{5}-1}{4}\)+\(\frac{\sqrt{5}+1}{4}\)
=\(\frac{\sqrt{5}-1+\sqrt{5}+1}{4}\)
=\(\frac{2\sqrt{5}}{4}\)
=\(\frac{\sqrt{5}}{2}\)

\(1)\dfrac{{14}}{{\sqrt 7 }} = \dfrac{{14\sqrt 7 }}{{\sqrt 7 .\sqrt 7 }} = \dfrac{{14\sqrt 7 }}{7} = 2\sqrt 7 \\ 2)\dfrac{{\sqrt 3 }}{{\sqrt 2 }} = \dfrac{{\sqrt 3 .\sqrt 2 }}{{\sqrt 2 .\sqrt 2 }} = \dfrac{{\sqrt 6 }}{2}\\ 3)\dfrac{5}{{\sqrt {10} }} = \dfrac{{5\sqrt {10} }}{{\sqrt {10} .\sqrt {10} }} = \dfrac{{5\sqrt {10} }}{{10}} = \dfrac{{\sqrt {10} }}{2}\\ 4)\dfrac{3}{{2\sqrt 5 }} = \dfrac{{3.2\sqrt 5 }}{{2\sqrt 5 .2\sqrt 5 }} = \dfrac{{6\sqrt 5 }}{{20}} = \dfrac{{3\sqrt 5 }}{{10}}\\ 5)\dfrac{{7 + \sqrt 7 }}{{\sqrt 7 + 1}} = \dfrac{{\left( {7 + \sqrt 7 } \right)\left( {\sqrt 7 - 1} \right)}}{{\left( {\sqrt 7 + 1} \right)\left( {\sqrt 7 - 1} \right)}} = \dfrac{{6\sqrt 7 }}{6} = \sqrt 7 \\ 6)\dfrac{{\sqrt 2 - \sqrt 6 }}{{3\sqrt 3 - 3}} = \dfrac{{\left( {\sqrt 2 - \sqrt 6 } \right)\left( {3\sqrt 3 + 3} \right)}}{{\left( {3\sqrt 3 - 3} \right)\left( {3\sqrt 3 + 3} \right)}} = \dfrac{{ - 2\sqrt 2 }}{6} = \dfrac{{ - \sqrt 2 }}{3}\\ 7)\dfrac{{\sqrt 3 }}{{3 - \sqrt 3 }} = \dfrac{{\sqrt 3 \left( {3 + \sqrt 3 } \right)}}{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}} = \dfrac{{3\sqrt 3 + 3}}{6} = \dfrac{{3\left( {\sqrt 3 + 1} \right)}}{6} = \dfrac{{\sqrt 3 + 1}}{2}\\ 8)\dfrac{2}{{2 - \sqrt 3 }} = \dfrac{{2\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 4 + 2\sqrt 3 \\ 9)\dfrac{{\sqrt 3 + 2}}{{2 - \sqrt 3 }} = \dfrac{{\left( {\sqrt 3 + 2} \right)\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 7 + 4\sqrt 3 \\ 10)\dfrac{{3\sqrt 5 }}{{2\sqrt 5 - 1}} = \dfrac{{3\sqrt 5 \left( {2\sqrt 5 + 1} \right)}}{{\left( {2\sqrt 5 - 1} \right)\left( {2\sqrt 5 + 1} \right)}} = \dfrac{{30 + 3\sqrt 5 }}{{19}}\\ 11)\dfrac{1}{{\sqrt 3 }} = \dfrac{{1.\sqrt 3 }}{{\sqrt 3 .\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3} \)

Bài rút gọn 

\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)

\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)

Bài gpt:

\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)

Đk:\(-1\le x\le3\)

\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)

Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm

Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)

d) \(\frac{1}{\sqrt{3}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}-\frac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{2\sqrt{5}}{3-5}=\frac{2\sqrt{5}}{-2}=-\sqrt{5}\)c) \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}+\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)

b) \(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}+\sqrt{5-2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)a) \(\sqrt{27}+\sqrt{243}-6\sqrt{12}=\sqrt{9.3}+\sqrt{81.3}-6\sqrt{4.3}=3\sqrt{3}+9\sqrt{3}-12\sqrt{3}=0\)