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a). \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)
\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\)
ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)
\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)
\(\Leftrightarrow C=-3\)
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
Câu A=4
Cách giải:
\(\left(5\sqrt{3}+2\sqrt{12}-\sqrt{75}\right):\sqrt{3}\)
\(=\left(5\sqrt{3}+2\sqrt{4\cdot3}-\sqrt{25\cdot3}\right)\)\(:\sqrt{3}\)
\(=\left(5\sqrt{3}+4\sqrt{3}-5\sqrt{3}\right)\)\(:\sqrt{3}\)
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/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
Trả lời
\(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
Đặt \(M=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)
\(M^2=\left(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\right)^2\)
\(M^2=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)^2}{\left(\sqrt{\sqrt{5}+1}\right)^2}\)
\(M^2=\frac{\sqrt{5}+2+2\sqrt{\left(\sqrt{5}+2\right).\left(\sqrt{5}-2\right)}+\sqrt{5}-2}{\sqrt{5}+1}\)
\(M^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}\)
\(M^2=\frac{2\sqrt{5}+2}{\sqrt{5}+1}\)
\(M^2=\frac{2.\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\)
\(M^2=2\)
\(M=\sqrt{2}\)
THay M vào B ta có \(B=M-\sqrt{3-2\sqrt{2}}\)
\(B=\sqrt{2}-\sqrt{3-2\sqrt{2}}\)
\(B=\sqrt{2}-\sqrt{2-2\sqrt{2}+1}\)
\(B=\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(B=\sqrt{2}-\sqrt{2}+1\)
\(B=1\)