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Nguyễn Minh Anh
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Tiểu Ma Bạc Hà
14 tháng 7 2019 lúc 22:34

\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)

\(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

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\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

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\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

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nood
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wary reus
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Trần Việt Linh
28 tháng 7 2016 lúc 10:15

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)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)

\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)

\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)

\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\sqrt{3}-1+\sqrt{3+1}\)

\(=2\sqrt{3}\)

 

 

 

 

Lương Ngọc Anh
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Nguyễn Lê Phước Thịnh
24 tháng 6 2023 lúc 20:14

g: \(=\left|\sqrt{6}-1\right|=\sqrt{6}-1\)

h: \(=\left|2\sqrt{3}-1\right|=2\sqrt{3}-1\)

l: \(=\left|2-\sqrt{3}\right|-2=2-\sqrt{3}-2=-\sqrt{3}\)

j: \(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)

\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

Phương Anh Nguyễn Thị
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Do What You Love
1 tháng 7 2017 lúc 8:27

a,\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\\ =\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}=5+3\sqrt{2}\)

Trần Anh Tuấn
1 tháng 7 2017 lúc 10:13

b, \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}+\sqrt{3+\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}\)

\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}\)

\(\Leftrightarrow\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(\Leftrightarrow\sqrt{3}-1+\sqrt{3}+1\)

\(\Leftrightarrow2\sqrt{3}\)

Trần Anh Tuấn
1 tháng 7 2017 lúc 10:41

d,\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2.\sqrt{2}\sqrt{3}+2}\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

\(\Leftrightarrow3-2\Leftrightarrow1\)

DŨNG NGUYỄN HACKER
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T=4,06731601

Nguyễn Linh Chi
10 tháng 12 2019 lúc 21:56

Em tham khảo đề bài và bài làm tại link: Câu hỏi của Trân Vũ Mai Ngọc - Toán lớp 9 - Học toán với OnlineMath

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Gia Bảo Hà Đình
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Akai Haruma
11 tháng 8 2021 lúc 17:44

Lời giải:
Gọi biểu thức là $A$

\(A=\frac{-\sqrt{3}(1-\sqrt{2})}{1-\sqrt{2}}+\frac{\sqrt{3}(\sqrt{3}+6)}{\sqrt{3}}-\frac{13(\sqrt{3}+4)}{(\sqrt{3}+4)(\sqrt{3}-4)}\)

\(=-\sqrt{3}+\sqrt{3}+6-\frac{13(\sqrt{3}+4)}{3-16}=6-(-\sqrt{3}-4)=10+\sqrt{3}\)

Phạm Hà Linh
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HT.Phong (9A5)
22 tháng 8 2023 lúc 14:17

\(A=\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}\)

\(A=\left(4+\sqrt{3}\right)\sqrt{4^2-2\cdot4\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(A=\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}\)

\(A=\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)\)

\(A=4^2-3\)

\(A=13\)

\(B=\dfrac{3}{4+\sqrt{13}}+\dfrac{\sqrt{52}}{2}-3\)

\(B=\dfrac{3\left(4-\sqrt{13}\right)}{\left(4-\sqrt{13}\right)\left(4+\sqrt{13}\right)}+\dfrac{2\sqrt{13}}{2}-3\)

\(B=\dfrac{3\left(4-\sqrt{13}\right)}{16-13}+\sqrt{13}-3\)

\(B=4-\sqrt{13}+\sqrt{13}-3\)

\(B=4-3\)

\(B=1\)

肖一战(Nick phụ)
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Nguyen Khanh Huyen
7 tháng 8 2018 lúc 9:28

\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

\(=\sqrt{1}=1\)

b,c

\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)

=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)

\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)

\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

Nguyễn Ngọc Nhã Hân
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Aki Tsuki
30 tháng 6 2018 lúc 12:58

a/ \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{1+2\cdot1\cdot2\sqrt{2}+8}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(1+2\sqrt{2}\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{25+2\cdot5\cdot3\sqrt{2}+18}=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)

Aki Tsuki
30 tháng 6 2018 lúc 13:07

b/ \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{3\left(5+2\sqrt{6}\right)}-\sqrt{2\left(5+2\sqrt{6}\right)}\)

\(=\sqrt{15+6\sqrt{6}}-\sqrt{10+4\sqrt{6}}\)

\(=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(2+\sqrt{6}\right)^2}\)

\(=3+\sqrt{6}-2-\sqrt{6}=1\)

c/ \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(=\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}+\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}\)

\(=\sqrt{5-1-2\sqrt{3}}+\sqrt{3+1+2\sqrt{3}}\)

\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)

\(=\sqrt{3}-1+1+\sqrt{3}=2\sqrt{3}\)

Aki Tsuki
30 tháng 6 2018 lúc 13:16

d/ \(\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)

\(=\sqrt{1+\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}+\sqrt{1-\sqrt{3-\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)

\(=\sqrt{1+\sqrt{4+2\sqrt{3}}}+\sqrt{1-\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{1+\sqrt{\left(\sqrt{3}+1\right)^2}}+\sqrt{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{1+\sqrt{3}+1}+\sqrt{1-\sqrt{3}+1}\)

\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)