\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
giúp với
K
Khách
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K
1
DH
3 tháng 9 2019
Tham khảo:
Câu hỏi của Thẩm Thiên Tình - Toán lớp 9 | Học trực tuyến
NN
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
1
Y
7 tháng 7 2023
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{4^2-2.4.\sqrt{2}+\sqrt{2^2}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\left|4-\sqrt{2}\right|}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left|\sqrt{3}-1\right|}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{3}-2}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\sqrt{3^2}-1^2\\ =3-1\\ =2\)
TH
1
2
31 tháng 8 2019
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{3}+1}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{2-\sqrt{3}}}\)
\(B=\sqrt{6+2\cdot\sqrt{4-2\sqrt{3}}}\)
\(B=\sqrt{6+2\cdot\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(B=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(B=\sqrt{4+2\sqrt{3}}\)
\(B=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(B=\sqrt{3}+1\)
NQ
0
NT
1
LC
4
2 tháng 10 2016
Ta có \(\sqrt{18-\sqrt{128}}\)
= \(\sqrt{18-8\sqrt{2}}\)
= \(\sqrt{16-2×4×\sqrt{2}+2}\)
= \(4-\sqrt{2}\)
Từ đó cái ban đầu
= \(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
= \(\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
= \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
= \(\sqrt{6+2\sqrt{3}-2}\)
= \(\sqrt{4+2\sqrt{3}}\)
= \(\sqrt{3}+1\)
DH
0
TT
1
Q
22 tháng 6 2017
\(A=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12+\sqrt{18}+\sqrt{128}}}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\left(3-\sqrt{\sqrt{2}+\sqrt{12+\sqrt{18}+\sqrt{128}}}\right)}}\)
\(=\sqrt{6+2\sqrt{2\left(3-\sqrt{\sqrt{2}+\sqrt{12+3\sqrt{2}+8\sqrt{2}}}\right)}}\)
\(=\sqrt{6+2\sqrt{2\left(3-\sqrt{\sqrt{2}+\sqrt{12+11\sqrt{2}}}\right)}}\)
\(=\sqrt{6+2\sqrt{6}-2\sqrt{\sqrt{2}+\sqrt{12+11\sqrt{2}}}}\)
Ta có: \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
= \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2\cdot\sqrt{2}\cdot\sqrt{16}+2}}}\)
=\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(\sqrt{16}-\sqrt{2}\right)^2}}}\)
=\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16}-\sqrt{2}}}\)
=\(\sqrt{6-2\sqrt{4+\sqrt{12}}}\)
=\(\sqrt{6-2\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}\)
=\(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}\cdot1+1}=\sqrt{\left(\sqrt{3}-1\right)^2}\)=\(\sqrt{3}-1\)
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(16-\sqrt{2}\right)^2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2+\sqrt{12}+16-\sqrt{2}}}}\)
\(=\sqrt{6-2\sqrt{16+\sqrt{12}}}\) \(=\sqrt{6-2\sqrt{16+2\sqrt{3}}}\)
Thật xin lỗi! Phân tích đến đây là mk tịt r! Bn đok qa có khi lại nghĩ ra đấy!