$\frac{3}{\sqrt{3}-2}+\frac{2}{2+\sqrt{3}}$
Những câu hỏi liên quan
Rút gọn biểu thức 1) frac{sqrt{5+2sqrt{6}}+sqrt{8+2sqrt{15}}}{sqrt{7+2sqrt{10}}}2) left(2+frac{3+sqrt{3}}{sqrt{3}+1}right)left(2+frac{3-sqrt{3}}{sqrt{3}-1}right):left(sqrt{5}-2right)3) left(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right).left(sqrt{6}+11right)4) frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+...+frac{1}{sqrt{99}+sqrt{100}}5) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-...-frac{1}{sqrt{98}-sqrt{99}}+frac{1}{sqrt{99}-s...
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Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
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Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
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Sao làm hổng ai bảo đú.n/g vậy :(((
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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}}
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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) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
A)\(\frac{6+2\sqrt{5}}{3-\sqrt{5}}-\frac{5+3\sqrt{5}}{\sqrt{5}}+\frac{\sqrt{5}}{2-\sqrt{5}}\)
B)\(\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}-\frac{3}{\sqrt{2}-1}\)
C)\(\frac{3+\sqrt{2}}{3-\sqrt{3}}-\frac{3+\sqrt{3}}{\sqrt{3}}-\frac{2}{\sqrt{3}-1}\)
D
a)frac{1}{sqrt{8}+sqrt{7}}+sqrt{175}-frac{6sqrt{2}-4}{3-sqrt{2}}
b)sqrt{2-sqrt{3}}-sqrt{frac{3}{2}}
c)frac{sqrt{30}-sqrt{2}}{sqrt{8-sqrt{15}}}-sqrt{8-sqrt{49+8sqrt{3}}}
d) frac{2+sqrt{3}}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{2-sqrt{3}}{sqrt{2}-sqrt{2-sqrt{3}}}
e)frac{sqrt{2}+sqrt{3}+sqrt{6}+sqrt{8}+4}{sqrt{2}+sqrt{3}+sqrt{4}}
f)frac{left(5+2sqrt{6}right)left(49-20sqrt{6}right)sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}
g)frac{frac{sqrt{2+sqrt{3}}}{2}}{frac{sqrt{2+sqrt{3}}}{2}-frac{2}{sqrt{6}}+frac{sqrt...
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a)\(\frac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\frac{6\sqrt{2}-4}{3-\sqrt{2}}\)
b)\(\sqrt{2-\sqrt{3}}-\sqrt{\frac{3}{2}}\)
c)\(\frac{\sqrt{30}-\sqrt{2}}{\sqrt{8-\sqrt{15}}}-\sqrt{8-\sqrt{49+8\sqrt{3}}}\)
d) \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
e)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
f)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
g)\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
C/M đẳng thức
a, \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
b, \(\frac{\sqrt{1+\frac{2\sqrt{2}}{3}}+\sqrt{1-\frac{2\sqrt{2}}{3}}}{\sqrt{1+\frac{2\sqrt{2}}{3}}-\sqrt{1-\frac{2\sqrt{2}}{3}}}=\sqrt{2}\)
Rút gọn biểu thức :
a,frac{2+sqrt{3}}{2-sqrt{3}};frac{5+2sqrt{6}}{5-2sqrt{6}}
b,frac{sqrt{3}-1}{sqrt{3}+1}
c,frac{2+sqrt{3}}{2-sqrt{3}}+frac{2-sqrt{3}}{2+sqrt{3}}
d,frac{sqrt{2+sqrt{3}}+sqrt{2-sqrt{3}}}{sqrt{2+sqrt{3}}-sqrt{2-sqrt{3}}}-frac{sqrt{2+sqrt{3}}-sqrt{2-sqrt{3}}}{sqrt{2+sqrt{3}}+sqrt{2-sqrt{3}}}
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Rút gọn biểu thức :
a,\(\frac{2+\sqrt{3}}{2-\sqrt{3}};\frac{5+2\sqrt{6}}{5-2\sqrt{6}}\)
b,\(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)
c,\(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
d,\(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}\)
a) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{\left(2+\sqrt{3}\right)^2}{4-3}\)
\(=\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=\frac{\left(5+2\sqrt{6}\right)^2}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\frac{\left(5+2\sqrt{6}\right)^2}{25-24}\)
\(=\left(5+2\sqrt{6}\right)^2=49+20\sqrt{6}\)
b) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3-2\sqrt{3}+1}{3-1}\)
\(=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
c) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}=14\)
d) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}\)
\(=\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}-\left(2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\right)}{2+\sqrt{3}-\left(2-\sqrt{3}\right)}\)
\(=\frac{4\sqrt{4-3}}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\)
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Tính: a)frac{3sqrt{2}-sqrt{6}}{3-sqrt{3}}+sqrt{frac{2-sqrt{2}}{2+sqrt{2}}}b) 6sqrt{frac{1}{3}}-frac{sqrt{3}-3}{sqrt{3}}+frac{3sqrt{2}-2sqrt{3}}{sqrt{2}-sqrt{3}}c) frac{3+sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-2sqrt{2}d) left(frac{sqrt{14}-sqrt{7}}{sqrt{2}-1}+frac{sqrt{15}-sqrt{5}}{sqrt{3}-1}right):frac{1}{sqrt{7}-sqrt{5}}Xin giúp 4 bài trên. Cảm ơn trước ạ!
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Tính: a)\(\frac{3\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}+\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}\)
b) \(6\sqrt{\frac{1}{3}}-\frac{\sqrt{3}-3}{\sqrt{3}}+\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{2}-\sqrt{3}}\)
c) \(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-2\sqrt{2}\)
d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
Xin giúp 4 bài trên. Cảm ơn trước ạ!
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
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b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
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Giúp mình với, ko cần làm hết đâu. Tính!a)frac{2+sqrt{3}}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{2-sqrt{3}}{sqrt{2}-sqrt{2-sqrt{3}}}b)frac{sqrt{2}}{2sqrt{2}+sqrt{3+sqrt{5}}}+frac{sqrt{2}}{2sqrt{2}-sqrt{3-sqrt{5}}}c)frac{left(5+2sqrt{6}right)left(49-20sqrt{6}right)sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}d)frac{frac{sqrt{2+sqrt{3}}}{2}}{frac{sqrt{2+sqrt{3}}}{2}-frac{2}{sqrt{6}}+frac{sqrt{2+sqrt{3}}}{2sqrt{3}}}
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Giúp mình với, ko cần làm hết đâu. Tính!
a)\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b)\(\frac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
c)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
d)\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\) =\(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\) =\(\frac{6}{6}=1\)
\(\Rightarrow A=\sqrt{2}\)
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Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
Thực hiện phép tính:
a)\(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\)
b) \(\frac{2}{3\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{3}+3\sqrt{3}}\)
c)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
d)\(\frac{3}{2\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{2}+3\sqrt{3}}\)
a) \(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{\left(3-\sqrt{2}\right)+\left(3+\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{6}{3^2-\left(\sqrt{2}\right)^2}=\frac{6}{7}\)
b) \(\frac{2}{3\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{3}+3\sqrt{3}}=\frac{2\left(2\sqrt{3}+3\sqrt{3}\right)-3\left(3\sqrt{2}-3\sqrt{3}\right)}{\left(3\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{3}+3\sqrt{3}\right)}=\frac{19\sqrt{3}-9\sqrt{2}}{-45+15\sqrt{6}}=-\frac{13\sqrt{3}+10\sqrt{2}}{15}\)c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{2}=\frac{16}{2}=8\)d) \(\frac{3}{2\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{2}+3\sqrt{3}}=\frac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=-\frac{18\sqrt{3}}{19}\)