Tính: \(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
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Thực hiện phép tính :
A = \(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
1/ frac{2}{3-sqrt{7}}sqrt{frac{6sqrt{2}-2sqrt{14}}{3sqrt{2}+sqrt{14}}}2/ sqrt{6+2sqrt{sqrt{5}-sqrt{13-sqrt{48}}}}3/ frac{sqrt{3-2sqrt{2}}}{sqrt{17-12sqrt{2}}}-frac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}4/ frac{24}{sqrt{7}+1}+frac{4}{3+sqrt{7}}-frac{3}{sqrt{7}+2}left(4-sqrt{7}right)5/ sqrt{7-3sqrt{5}}left(7+3sqrt{5}right)left(3sqrt{2}+sqrt{10}right)
Đọc tiếp
1/ \(\frac{2}{3-\sqrt{7}}\sqrt{\frac{6\sqrt{2}-2\sqrt{14}}{3\sqrt{2}+\sqrt{14}}}\)
2/ \(\sqrt{6+2\sqrt{\sqrt{5}-\sqrt{13-\sqrt{48}}}}\)
3/ \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
4/ \(\frac{24}{\sqrt{7}+1}+\frac{4}{3+\sqrt{7}}-\frac{3}{\sqrt{7}+2}\left(4-\sqrt{7}\right)\)
5/ \(\sqrt{7-3\sqrt{5}}\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
Thực hiện phép tính:
\(A=\dfrac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
Dễ dàng nhận thấy \(\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}>0\)
Đặt \(a=\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}>0\)
\(\Rightarrow a^2=10-2\sqrt{25-17}=10-2\sqrt{8}=10-4\sqrt{2}\)
\(\Rightarrow a=\sqrt{10-4\sqrt{2}}\) (do \(a>0\) )
Đặt \(b=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\), tương tự dễ dàng c/m \(b>0\)
\(\Rightarrow b^2=6-2\sqrt{9-5}=2\Rightarrow b=\sqrt{2}\)
Vậy \(A=\dfrac{a-\sqrt{10-4\sqrt{2}}+4}{b+2-\sqrt{2}}=\dfrac{\sqrt{10-4\sqrt{2}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{2}+2-\sqrt{2}}\)
\(\Rightarrow A=\dfrac{4}{2}=2\)
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Tinh
a, \(3\sqrt{50}-2\sqrt{98}-5\sqrt{18}-\sqrt{63}-2\sqrt{28}\)
b, \(\sqrt{42-10\sqrt{17}}+\sqrt{3-8\sqrt{17}}\)
c, \(\frac{4}{\sqrt{3}+1}+\frac{6}{\sqrt{3}-3}-\frac{5}{\sqrt{3}-2}\)
Bài 1: Tính
A=\(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
Bài 2: Tìm x,y nguyên thỏa mãn:
y2 = 1 + \(\sqrt{9-x^2-4x}\)
Help me!!! Giúp với... Mình gần thi học sinh giỏi rồi.... Có ai wan tâm ko? Huhuhu...
a) \(\dfrac{5-2\sqrt{ }5}{\sqrt{ }5-2}-\dfrac{11}{4+\sqrt{ }5} \)
b)\(\sqrt{9+4\sqrt{ }5-\sqrt{ }6-2\sqrt{ }5}\)
c)\(\sqrt{17-3\sqrt{ }32+\sqrt{ }17+\sqrt{ }32}\)
\(\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{11\left(4-\sqrt{5}\right)}{16-5}=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)
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\(=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)
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1. sqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}+sqrt{8}
2. dfrac{sqrt{3-2sqrt{3}}}{sqrt{17-12sqrt{2}}}-dfrac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
3.sqrt{7+2sqrt{6}}-sqrt{left(sqrt{6-1}right)^2}
4sqrt{5-2sqrt{6}}-sqrt{5+sqrt{24}}
5.sqrt{4sqrt{5+sqrt{3+5sqrt{48-10sqrt{7+4sqrt{3}}}}}}
6.sqrt{6+2sqrt{5-sqrt{13+sqrt{48}}}}
Đọc tiếp
1. \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\)
2. \(\dfrac{\sqrt{3-2\sqrt{3}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
3.\(\sqrt{7+2\sqrt{6}}-\sqrt{\left(\sqrt{6-1}\right)^2}\)
4\(\sqrt{5-2\sqrt{6}}-\sqrt{5+\sqrt{24}}\)
5.\(\sqrt{4\sqrt{5+\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
6.\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
1. sqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}+sqrt{8}
2. dfrac{sqrt{3-2sqrt{3}}}{sqrt{17-12sqrt{2}}}-dfrac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
3.sqrt{7+2sqrt{6}}-sqrt{left(sqrt{6}-1right)^2}
4sqrt{5-2sqrt{6}}-sqrt{5+sqrt{24}}
5.sqrt{4sqrt{5+sqrt{3+5sqrt{48-10sqrt{7+4sqrt{3}}}}}}
6.sqrt{6+2sqrt{5-sqrt{13+sqrt{48}}}}
Đọc tiếp
1. \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\)
2. \(\dfrac{\sqrt{3-2\sqrt{3}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
3.\(\sqrt{7+2\sqrt{6}}-\sqrt{\left(\sqrt{6}-1\right)^2}\)
4\(\sqrt{5-2\sqrt{6}}-\sqrt{5+\sqrt{24}}\)
5.\(\sqrt{4\sqrt{5+\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
6.\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
Tính
1/ \(\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
2/ \(\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
3/ \(\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
4/ \(\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}\)
\(1.\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\sqrt{17-4\sqrt{5+2.2\sqrt{5}+4}}=\sqrt{17-4\left(\sqrt{5}+2\right)}=\sqrt{5-2.2\sqrt{5}+4}=\sqrt{5}-2\)
\(2.\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{17-6\sqrt{2+2\sqrt{2}+1}}=\sqrt{17-6\left(\sqrt{2}+1\right)}=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)\(3.\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}=\sqrt{3+\sqrt{3-2\sqrt{3}+1}}=\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(4.\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}=\sqrt{25+2.5\sqrt{2}+2}.\left(5-\sqrt{2}\right)=\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)=5-2=3\)
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