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\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\\ =\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}\\ =\sqrt{1}=1\)
\(B=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{3+1-2\sqrt{3}}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\sqrt{3}-2}\\ =\sqrt{3+1+2\sqrt{3}}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-10\sqrt{3}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\\ =\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)
\(D=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\\ \text{Ta có }:\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)^2\\ =3+\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+3-\sqrt{5}\\ =6-2\sqrt{9-5}=6-2\sqrt{4}=6-4=2\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{2}\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
a) \(\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{29-6\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-2\sqrt{5}+3}\)
\(=\sqrt{3-\sqrt{3}-\sqrt{5}}\)
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
~ ~ ~ ~ ~
\(M=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\sqrt{3}-2}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}+1\)
7.
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+2\sqrt{4.3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{(\sqrt{4}+\sqrt{3})^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(2+\sqrt{3})}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-2.5\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5(5-\sqrt{3})}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
5.
\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)
6.
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)
\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{(\sqrt{2}+\sqrt{5})^2}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}-\sqrt{(\sqrt{2}+\sqrt{5})^2}=|\sqrt{2}+\sqrt{5}+1|-|\sqrt{2}+\sqrt{5}|=1\)
a/ A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
Vậy A = 1
b) \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}+\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}-\sqrt{2}.\sqrt{6-2\sqrt{5}}\)
\(=\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}+\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}-\sqrt{2}.\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\sqrt{2}.\left(\sqrt{5}-1\right)\)
\(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)+\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}}-\sqrt{10}+\sqrt{2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\sqrt{10}+\sqrt{2}=\dfrac{2\sqrt{5}}{\sqrt{2}}-\sqrt{10}+\sqrt{2}\)
\(=\sqrt{10}-\sqrt{10}+\sqrt{2}=\sqrt{2}\)
e) \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(C=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(C=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(C=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
câu a ; f chưa nghỉ ra
Câu a:
\( \sqrt {4 + \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {7 + 4\sqrt 3 } } } } \\ = \sqrt {4 + \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} } } } \\ = \sqrt {4 + \sqrt {5\sqrt 3 + 5\sqrt {48 - 20 - 10\sqrt 3 } } } \\ = \sqrt {4 + \sqrt {5\sqrt 3 + 5\sqrt {28 - 10\sqrt 3 } } } \\ = \sqrt {4 + \sqrt {5\sqrt 3 + 5\sqrt {{{\left( {5 - \sqrt 3 } \right)}^2}} } } \\ = \sqrt {4 + \sqrt {5\sqrt 3 + 25 - 5\sqrt 3 } } \\ = \sqrt {4 + \sqrt {25} } = \sqrt {4 + 5} = \sqrt 9 = 3 \)
Câu b:
\( \sqrt {\sqrt {5 - \sqrt {3 - \sqrt {29 - 12\sqrt 5 } } } } \\ = \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {{{\left( {2\sqrt 5 } \right)}^2} - 2.2\sqrt 5 .3 + {3^2}} } } \\ = \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {{{\left( {2\sqrt 5 - 3} \right)}^2}} } } \\ = \sqrt {\sqrt 5 - \sqrt {3 - \left| {2\sqrt 5 - 3} \right|} } \\ = \sqrt {\sqrt 5 - \sqrt {3 - \left( {2\sqrt 5 - 3} \right)} } \\ = \sqrt {\sqrt 5 - \sqrt {3 - 2\sqrt 5 + 3} } \\ = \sqrt {\sqrt 5 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} } \\ = \sqrt {\sqrt 5 - \sqrt 5 + 1} = \sqrt 1 = 1 \)