Tính
\(A=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12+\sqrt{18}-\sqrt{128}}}}}\)
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TT
1
Những câu hỏi liên quan
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
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{4^2-2.4.\sqrt{2}+\sqrt{2^2}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\left|4-\sqrt{2}\right|}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left|\sqrt{3}-1\right|}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{3}-2}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\sqrt{3^2}-1^2\\ =3-1\\ =2\)
NN
2
K
1
DH
3 tháng 9 2019
Tham khảo:
Câu hỏi của Thẩm Thiên Tình - Toán lớp 9 | Học trực tuyến
TH
1
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
\(A=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\left(4-\sqrt{2}\right)}}}\)
\(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{4+2\sqrt{3}}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)^2=\sqrt{2}+4+2\sqrt{3}\)
28 tháng 10 2022
\(=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
MT
15 tháng 8 2015
\(A=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-2.4.\sqrt{2}+2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}\)
\(=\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{6-2\sqrt{3+2\sqrt{3}.1+1}}\)
\(=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{6-2\sqrt{3}-2}\)
\(=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}.1+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
15 tháng 8 2015
b) B = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)
= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
2
31 tháng 8 2019
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{3}+1}}\)
\(B=\sqrt{6+2\sqrt{2}\cdot\sqrt{2-\sqrt{3}}}\)
\(B=\sqrt{6+2\cdot\sqrt{4-2\sqrt{3}}}\)
\(B=\sqrt{6+2\cdot\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(B=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(B=\sqrt{4+2\sqrt{3}}\)
\(B=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(B=\sqrt{3}+1\)
\(A=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12+\sqrt{18}+\sqrt{128}}}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\left(3-\sqrt{\sqrt{2}+\sqrt{12+\sqrt{18}+\sqrt{128}}}\right)}}\)
\(=\sqrt{6+2\sqrt{2\left(3-\sqrt{\sqrt{2}+\sqrt{12+3\sqrt{2}+8\sqrt{2}}}\right)}}\)
\(=\sqrt{6+2\sqrt{2\left(3-\sqrt{\sqrt{2}+\sqrt{12+11\sqrt{2}}}\right)}}\)
\(=\sqrt{6+2\sqrt{6}-2\sqrt{\sqrt{2}+\sqrt{12+11\sqrt{2}}}}\)
-128 chứ đou phải +128