tính: \((\dfrac{\sqrt{4}}{3}-\sqrt{3}+\dfrac{\sqrt{25}}{3})\cdot\sqrt{12}\)
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
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15 tháng 12 2021 lúc 21:58
Thực hiện phép tính (tính nhanh nếu có thể):
4) \(4\cdot\left(\dfrac{-1}{2}\right)^3+\left|-1\dfrac{1}{2}+\sqrt{\dfrac{9}{4}}\right|:\sqrt{25}\)
5) \(\left[6-3\cdot\left(\dfrac{-1}{3}\right)^2+\sqrt{\dfrac{1}{4}}\right]:\sqrt{0,\left(9\right)}\)
D sqrt{94-42sqrt{5}}-sqrt{94+42sqrt{5}}
A sqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
C dfrac{2sqrt{4-sqrt{5+sqrt{21+sqrt{80}}}}}{sqrt{10}-sqrt{2}}
F sqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{3}}}}
B dfrac{1}{sqrt{1}+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+...+dfrac{1}{sqrt{n-1}+sqrt{n}}
E dfrac{1}{sqrt{1}-sqrt{2}}-dfrac{1}{sqrt{2}-sqrt{3}}+dfrac{1}{sqrt{3}-sqrt{4}}-...-dfrac{1}{sqrt{24}-sqrt{25}}
Đọc tiếp
D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
F = \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
B = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}\)
E = \(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{24}-\sqrt{25}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{6+\sqrt{20}}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)}{10-2}\)
C = \(\dfrac{2\sqrt{30-10\sqrt{5}}+2\sqrt{6-2\sqrt{5}}}{8}\)
C = \(\dfrac{2\sqrt{\left(5-\sqrt{5}\right)^2}+2\sqrt{\left(\sqrt{5}-1\right)^2}}{8}\)
C = \(\dfrac{2\left(5-\sqrt{5}\right)+2\left(\sqrt{5}-1\right)}{8}\)
C = \(\dfrac{10-2\sqrt{5}+2\sqrt{5}-2}{8}\) = \(\dfrac{8}{8}\) = \(1\)
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D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
D = \(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)
D = \(7-3\sqrt{5}-\left(7+3\sqrt{5}\right)\) = \(7-3\sqrt{5}-7-3\sqrt{5}\)
D = \(-6\sqrt{5}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\) = \(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{5}+1}\) = \(\sqrt{1}=1\)
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Tính:
\(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
Ta có: \(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\left(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{6}+2\right)}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\left(5+\sqrt{6}+\sqrt{6}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\dfrac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}{4}\)
\(=\dfrac{25-24}{4}=\dfrac{1}{4}\)
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\(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4\cdot2-4\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4}\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\)
\(\dfrac{3}{5}+\dfrac{\left(-1\right)}{4},\dfrac{|-3|}{5}\cdot\dfrac{4}{9}+\dfrac{\left(-4\right)}{9},\sqrt{16}+\sqrt{25}\)
1, dfrac{6-sqrt{6}}{sqrt{6}-1}+dfrac{6+sqrt{6}}{sqrt{6}}
2, dfrac{6-6sqrt{3}}{1-sqrt{3}}+dfrac{3sqrt{3}+3}{sqrt{3}+1}
3, dfrac{3+sqrt{3}}{sqrt{3}}+dfrac{sqrt{6}-sqrt{3}}{1-sqrt{2}}
4, dfrac{sqrt{15}-sqrt{12}}{sqrt{5}-2}+dfrac{6+2sqrt{6}}{sqrt{3}+sqrt{2}}
5, left(dfrac{3sqrt{125}}{15}-dfrac{10-4sqrt{5}}{sqrt{5}-2}right)cdotdfrac{1}{sqrt{5}}
Đọc tiếp
1, \(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6+\sqrt{6}}{\sqrt{6}}\)
2, \(\dfrac{6-6\sqrt{3}}{1-\sqrt{3}}+\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
3, \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
4, \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
5, \(\left(\dfrac{3\sqrt{125}}{15}-\dfrac{10-4\sqrt{5}}{\sqrt{5}-2}\right)\cdot\dfrac{1}{\sqrt{5}}\)
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
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thực hiện phép tính
A=\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
B=\(\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{2}}}\cdot\left(3+\sqrt{5}\right)\)
a) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\dfrac{2-\sqrt{3}}{1}+\dfrac{2+\sqrt{3}}{1}\)
=4
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1. Tính : \(\dfrac{12}{4-\sqrt{10}}\)-6\(\sqrt{\dfrac{5}{2}}\)+\(\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=(\(\dfrac{\sqrt{x}}{\sqrt{x}-5}\)-\(\dfrac{5}{\sqrt{x}+5}\)+\(\dfrac{10\sqrt{x}}{25-x}\)):\(\dfrac{3}{\sqrt{x}+5}\)
1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)
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1,Tính \(\dfrac{12}{4-\sqrt{10}}-6\sqrt{\dfrac{5}{2}}+\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{5}{\sqrt{x}+5}+\dfrac{10\sqrt{x}}{25-x}\right):\dfrac{3}{\sqrt{x}+5}\)
1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)
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