Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Trục căn thức ở mẫu và rút gọn
a, (\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\))(\(\sqrt{6} +11\))
b,(\(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\))(\(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\))
c,\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\)- (\(\sqrt{2}+\sqrt{3}\))
d,(\(\frac{5-2\sqrt{5}}{2-\sqrt{5}}-2\))(\(\frac{5+3\sqrt{5}}{3+\sqrt{5}}-2\))
giúp mik vs
trục căn thức ở mẫu
\(\frac{1\sqrt{2}}{\sqrt{2}.\sqrt{2}}\) \(\frac{1\sqrt[]{3}}{\sqrt{3}.\sqrt{3}}\)\(\frac{2\sqrt{5}-5}{\sqrt{5}-2}\)\(\frac{6-2\sqrt{6}}{2-\sqrt{6}}\)\(\frac{5\sqrt{6}-6\sqrt{5}}{\sqrt{5}-\sqrt{6}}\)\(\frac{1}{\sqrt{5}-1}\)\(\frac{1}{4-2\sqrt{a}+a}\)\(\frac{1}{b-2\sqrt{b}+4}\)\(\frac{1}{b-3\sqrt{b}+9}\)
trục căn thức ở mẫu và thực hiện phép tính
a)\(\frac{2}{\sqrt{5}+2}+\frac{1}{\sqrt{3}-2}-2\sqrt{5}\)
b)\(\sqrt{\frac{2}{2-\sqrt{3}}}-\sqrt{\frac{2}{2+\sqrt{3}}}\)
c)\(\frac{2}{\sqrt{3}+1}-\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}+3}\)
d)\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{35}+\sqrt{36}}\)
Bài 1 Trục căn thức ở mẫu
a,\(\frac{26}{5-2\sqrt{3}}\)
b,\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\)
c,\(\frac{2\sqrt{10}-5}{4-\sqrt{10}}\)
d,\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
e,\(\frac{1}{\sqrt{5}-\sqrt{3}+2}\)
f,\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
a. \(\frac{26}{5-2\sqrt{3}}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}=\frac{26\cdot\left(5+2\sqrt{3}\right)}{13}=2\cdot\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)
b.\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\sqrt{3}\cdot\left(3\sqrt{3}-2\right)}{\sqrt{2}\cdot\left(3\sqrt{3}-2\right)}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}\)
c.\(\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\sqrt{5}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{10}}{2}\)
d.\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)=\(4\sqrt{5}\)
Rút gọn :
\(B=\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(C=\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
\(D=\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(E=\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
\(F=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
1. Tìm trục căn ở mẫu
a, \(\frac{3}{\sqrt{5}-\sqrt{2}}\) ; b,\(\frac{\sqrt{2}+1}{\sqrt{2}-1}\) ; c, \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\) ; d, \(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)
2. Rút gọn biểu thức:
A=\(\frac{1}{7+\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
B=\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
Trục căn thức ở mẫu:
a,\(\frac{1}{\sqrt{2}-1}\)
b,\(\frac{2}{\sqrt{3}+1}\)
c,\(\frac{5}{\sqrt{7}-\sqrt{2}}\)
d,\(\frac{6}{2\sqrt{3}+\sqrt{2}}\)
e,\(\frac{1}{2\sqrt{a}+1}\)
g,\(\frac{2xy}{2\sqrt{x}+3\sqrt{y}}\)
h,\(\frac{x\sqrt{x}-1}{\sqrt{x}-1}\)
i,\(\frac{a-9b}{\sqrt{a}-3\sqrt{b}}\)
k,\(\frac{15-2\sqrt{5}}{3\sqrt{15}-2\sqrt{3}}\)
a) \(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}+\frac{12}{\sqrt{6}-3}-\sqrt{6}\)b)\(\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{\sqrt{3}}{2-\sqrt{6}}+\frac{\sqrt{3}}{2+\sqrt{6}}\right)-\frac{1}{\sqrt{2}}\)c) \(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right)\frac{1}{\sqrt{3}+5}\)d) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
