Tính:
a.\(A=\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
b.\(B=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
c.\(C=\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
d.\(D=2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}\)
Giúp em với ạ!
Những câu hỏi liên quan
thực hiện phép tính
a, \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b,\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{3}}}}\)
c,\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
d,\(\sqrt{5-\sqrt{13+4\sqrt{3}}+}\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20-2\cdot3\cdot\sqrt{20}+9}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20}+3}}\)
\(=\sqrt{5-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{5-\sqrt{5-2\sqrt{5}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}-1}\)
\(=\sqrt{4-\sqrt{5}}\)
c)\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=3-2=1\)
d)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3+1}\)
\(=2\sqrt{3}\)
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Tính:
a.\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b.\(\sqrt{13+\sqrt{30\sqrt{2+\sqrt{9+4\sqrt{2}}}}}\)
c.\(\frac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}\)
a/ \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)
b,c tương tự
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bài 5 Tính:
a) \(\sqrt{6-2\sqrt{5}}\)
b)\(\sqrt{7-4\sqrt{3}}\)
c)\(\sqrt{3-2\sqrt{2}}\) -\(\sqrt{6-4\sqrt{2}}\)
d)\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
Lời giải:
a. \(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}.\sqrt{1}+1}=\sqrt{(\sqrt{5}-1)^2}=\sqrt{5}-1\)
b. \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\sqrt{4}.\sqrt{3}+3}=\sqrt{(\sqrt{4}-\sqrt{3})^2}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}\)
c.
\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)
\(=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{4}-\sqrt{2})^2}\)
\(=|\sqrt{2}-1|-|\sqrt{4}-\sqrt{2}|=\sqrt{2}-1-(2-\sqrt{2})=2\sqrt{2}-3\)
d.
\(=\sqrt{13+30\sqrt{2+\sqrt{(\sqrt{8}+1)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}\)
\(=\sqrt{13+30(\sqrt{2}+1)}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2\sqrt{18.25}+25}\)
\(=\sqrt{(\sqrt{18}+\sqrt{25})^2}=\sqrt{18}+\sqrt{25}=5+3\sqrt{2}\)
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a) \(\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)
b) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
c) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2}-1-2+\sqrt{2}=-3+2\sqrt{2}\)
d) Ta có: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=5+3\sqrt{2}\)
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Giúp mình với
II.nhân:\(\sqrt{A}\).\(\sqrt{B}\)=\(\sqrt{..............}\)(A≥0;B≥0)
a)\(\sqrt{2}\left(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\right)\)
b)\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
c)\(\sqrt{6+2\sqrt{5}-\sqrt{13+\sqrt{48}}}\)
a) Ta có: \(\sqrt{2}\left(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) Ta có: \(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{2}+2\sqrt{2}+1}\)
\(=\sqrt{14+32\sqrt{2}}\)
c) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{13+\sqrt{48}}}\)
\(=\sqrt{6+2\sqrt{5}-2\sqrt{3}-1}\)
\(=\sqrt{5+2\sqrt{5}-2\sqrt{3}}\)
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Tính:
a, A = \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
b, B = \(\sqrt{6-2\sqrt{5}+\sqrt{6+2\sqrt{5}}}\)
c, C = \(\sqrt{1+2\sqrt{1+2\sqrt{3+2\sqrt{2}}}}\)
d, D = \(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
e, E = \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
Giúp mình tính bài này với ạ =)) cần gấp.
a) \(\left(2+\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
b) \(\sqrt{6-2\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}\)
c) \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
d) \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}-3\sqrt{2}\)
c/ = \(\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{25+2.3.5.\sqrt{2}+18}\)
\(=5+3\sqrt{2}\)
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d/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
\(=\sqrt{13+6\left(\sqrt{3}+1\right)}\)
\(=\sqrt{19+6\sqrt{2}}\)
\(=3\sqrt{2}+1\)
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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= \(\dfrac{\left(5+4\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
https://hoc24.vn/hoi-dap/question/407636.html
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}\)
= 9
~ ~ ~ ~ ~
\(M=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\sqrt{3}-2}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}+1\)
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\(M=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
= 1
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1 .
a)Afrac{2}{2sqrt[3]{2}+2+sqrt[3]{4}}
b)Bfrac{6}{2sqrt[3]{2}-2+sqrt[3]{4}}
c)Cfrac{2}{sqrt[3]{4}+sqrt[3]{2}+2}
2 .
a)Asqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
b)Bfrac{left(5+2sqrt{6}right).left(49-20sqrt{6}right).sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}
c)Csqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7}+4sqrt{3}}}}
d)D(left(sqrt{3}-1right).sqrt{6+2sqrt{2}.sqrt{3-sqrt{sqrt{2}+sqrt{12}+sqrt{18-sqrt{128}}}}}
Đọc tiếp
1 .
a)\(A=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
b)\(B=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\)
c)C=\(\frac{2}{\sqrt[3]{4}+\sqrt[3]{2}+2}\)
2 .
a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b)\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
c)C=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7}+4\sqrt{3}}}}\)
d)D=(\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Tính \(C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
Tính\(D=\sqrt{6+2\sqrt{5-\sqrt{29-12\sqrt{5}}}}\)
a,C=\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
=\(\sqrt{13+30\sqrt{2+\sqrt{9+2\sqrt{8}}}}\)=\(\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)
=\(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
=\(\sqrt{13+30\sqrt{2}+30}=\sqrt{43+2.5.3\sqrt{2}}\)
=\(\sqrt{43+2.5.\sqrt{18}}=\sqrt{\left(\sqrt{18}+5\right)^2}=\sqrt{18}+5\)
=> C=\(\sqrt{18}+5\)
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Link nè bạn: (Tính D)
https://hoc24.vn/hoi-dap/question/401820.html
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