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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/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
a/ Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)
Khi đó ta có a2 + b2 = 6; ab = 2; a + b = \(\sqrt{10}\) ; a - b = \(\sqrt{2}\); a2 - b2 = \(2\sqrt{5}\)
Ta có cái ban đầu
\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)=
\(\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)
\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}=\frac{6\sqrt{2}}{11}\)
a, \(\sqrt{2}A=\sqrt{10-2\sqrt{3.7}}+\sqrt{10+2\sqrt{3.7}}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{3}+\sqrt{7}=2\sqrt{7}\)
\(\Rightarrow A=\sqrt{14}\)
b, \(B=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{3\sqrt{5}}{2}\)
c, \(C=\left(1-\sqrt{11}\right)\left(\sqrt{11}+1\right)=1-11=-10\)
d, \(D=\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}{2-3}-\frac{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}{2-3}\)
\(=-2-\sqrt{6}+2-\sqrt{6}=-2\sqrt{6}\)
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 \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
Xét biểu thức phụ : \(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng : \(\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+...+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
\(\frac{\sqrt{5}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{2\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)Rút gọn
=\(\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
=\(\frac{5-\sqrt{15}-2\sqrt{15}+6-10-2\sqrt{15}-\sqrt{15}-3}{5-3}\)
\(=\frac{-2-6\sqrt{15}}{2}=\frac{-2\left(1+3\sqrt{15}\right)}{3}=-1-3\sqrt{15}\)