\(A=\sqrt{47+\sqrt{5}}\cdot\sqrt{47-\sqrt{5}}\)
\(=\sqrt{2204}=2\sqrt{551}\)
\(B=5-2\sqrt{6}+10+\sqrt{6}=15-\sqrt{6}\)
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Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
Rút gọn
\(A=\sqrt{47+\sqrt{5}}.\sqrt{7-\sqrt{2+5}}.\sqrt{7-\sqrt{2+5}}\)
\(B=\sqrt{49-20\sqrt{6}}+\sqrt{106+20\sqrt{6}}\)
\(C=\sqrt{302-20\sqrt{6}}+\sqrt{203-20\sqrt{6}}\)
MÌNH CẦN GẤP
\(A=\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}+2\sqrt{2}\\ B=\left(5+2\sqrt{6}\right)\cdot\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(C=\dfrac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\dfrac{1}{4}\sqrt{120}-\sqrt{\dfrac{15}{2}}\)
\(D=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}+\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
Tính giá trị các biểu thức sau:
a) \(A=\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
b) \(A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
c) \(A=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Rút gọn \(A=\left(\sqrt{6+\sqrt{20}}-2\sqrt{3-\sqrt{5}}+\sqrt{15-10\sqrt{2}}\right):\left(2+\sqrt{8}\right)\)
Rút gọn
a) \(\sqrt{6+3\sqrt{3}+\sqrt{6-3\sqrt{3}}}\)
b) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
c) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
1.Chứng minh
a) \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)
b) A= \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\) là số nguyên.
Bài 1: Tính
1, Aleft(1-frac{5+sqrt{5}}{1+sqrt{5}}right).left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
2, Bleft(frac{3sqrt{125}}{15}-frac{10-4sqrt{6}}{sqrt{5}-2}right).frac{1}{sqrt{5}}
3, Cleft(frac{sqrt{1000}}{100}-frac{5sqrt{2}-2sqrt{5}}{2sqrt{5}-8}right).frac{sqrt{10}}{10}
4, Dfrac{1}{sqrt{49+20sqrt{6}}}-frac{1}{sqrt{49-20sqrt{6}}}+frac{1}{sqrt{7-4sqrt{3}}}
5, Efrac{1}{sqrt{4-2sqrt{3}}}-frac{1}{sqrt{7-sqrt{48}}}+frac{3}{sqrt{14-6sqrt{5}}}
6, Ffrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}...
Đọc tiếp
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, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Bài 1: Rút gọn
a)sqrt{4+sqrt{10+2sqrt{5}}}+sqrt{4-sqrt{10+2sqrt{5}}} ,
b)sqrt{4+sqrt{15}}+sqrt{4-sqrt{15}}-2sqrt{3-sqrt{5}}
c)Asqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}
d)Bsqrt{6+2sqrt{2}sqrt{3-sqrt{sqrt{2}+sqrt{12}+sqrt{18-8sqrt{2}}}}}
e)Csqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
f)D dfrac{left(5+4sqrt{6}right)left(49-20sqrt{6}right)left(5-2sqrt{6}right)sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}
Đọc tiếp
Bài 1: Rút gọn
a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,
b)\(\sqrt{4+\sqrt{15}}\)+\(\sqrt{4-\sqrt{15}}\)-\(2\sqrt{3-\sqrt{5}}\)
c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f)D= \(\dfrac{\left(5+4\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Bài 1: Rút gọn
a)sqrt{4+sqrt{10+2sqrt{5}}}+sqrt{4-sqrt{10+2sqrt{5}}} ,
b)sqrt{4+sqrt{15}}+sqrt{4-sqrt{15}}-2sqrt{3-sqrt{5}}
c)Asqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}
d)Bsqrt{6+2sqrt{2}sqrt{3-sqrt{sqrt{2}+sqrt{12}+sqrt{18-8sqrt{2}}}}}
e)Csqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
f)Ddfrac{left(5+2sqrt{6}right)left(49-20sqrt{6}right)left(5-2sqrt{6}right)sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}
Mình sửa lại để m.n dễ nhìn hơn!
Đọc tiếp
Bài 1: Rút gọn
a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,
b)\(\sqrt{4+\sqrt{15}}\)+\(\sqrt{4-\sqrt{15}}\)-\(2\sqrt{3-\sqrt{5}}\)
c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f)D=\(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Mình sửa lại để m.n dễ nhìn hơn!