giúp mình với
\(\sqrt{7-\sqrt{2\sqrt{2+\sqrt{50+\sqrt{18-\sqrt{128}}}}}}\)
K
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R
0
Những câu hỏi liên quan
NH
3
CW
1 tháng 8 2017
kiểm tra bằng máy tính:
\(2\sqrt{2+\sqrt{50}\sqrt{18-\sqrt{128}}}>7\)
căn thức ko có nghĩa
NM
1
27 tháng 6 2017
\(\sqrt{7-2\sqrt{2+5\sqrt{2}+\sqrt{18-2\cdot4\cdot\sqrt{2}}}}\)=\(\sqrt{7-2\sqrt{2+5\sqrt{2}+4-\sqrt{2}}}\)
=\(\sqrt{7-2\sqrt{6+4\sqrt{2}}}=\sqrt{7-2\left(2+\sqrt{2}\right)}\) =\(\sqrt{3+2\sqrt{2}}\) =\(\sqrt{2}+1\)
TM
1
24 tháng 9 2019
\(\sqrt{18-\sqrt{128}}=\sqrt{18-8\sqrt{2}}=\sqrt{16-2.4.\sqrt{2}+2}=\sqrt{\left(4-\sqrt{2}\right)^2}=4-\sqrt{2}\)
=> \(\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}=\sqrt{2+5\sqrt{2}+4-\sqrt{2}}=\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{4+2.2\sqrt{2}+2}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)
=> \(\sqrt{7-2\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{7-2\left(2+\sqrt{2}\right)}=\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}.1+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
NM
2
23 tháng 7 2016
Ta có: \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
= \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2\cdot\sqrt{2}\cdot\sqrt{16}+2}}}\)
=\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(\sqrt{16}-\sqrt{2}\right)^2}}}\)
=\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16}-\sqrt{2}}}\)
=\(\sqrt{6-2\sqrt{4+\sqrt{12}}}\)
=\(\sqrt{6-2\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}\)
=\(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}\cdot1+1}=\sqrt{\left(\sqrt{3}-1\right)^2}\)=\(\sqrt{3}-1\)
23 tháng 7 2016
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(16-\sqrt{2}\right)^2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2+\sqrt{12}+16-\sqrt{2}}}}\)
\(=\sqrt{6-2\sqrt{16+\sqrt{12}}}\) \(=\sqrt{6-2\sqrt{16+2\sqrt{3}}}\)
Thật xin lỗi! Phân tích đến đây là mk tịt r! Bn đok qa có khi lại nghĩ ra đấy!
KT
6 tháng 7 2018
\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)
\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+1\)
KT
6 tháng 7 2018
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
\(A=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\left(4-\sqrt{2}\right)}}}\)
\(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{4+2\sqrt{3}}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)^2=\sqrt{2}+4+2\sqrt{3}\)
Y
17 tháng 6 2019
\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+\sqrt{12}}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\frac{4+\sqrt{4^2-7}}{2}}+\sqrt{\frac{4-\sqrt{4^2-7}}{2}}-\left(\sqrt{\frac{4+\sqrt{4^2-7}}{2}}-\sqrt{\frac{4-\sqrt{4^2-7}}{2}}\right)+\left(\sqrt{3}+1\right)^2\)
( áp dụng công thức căn phức tạp )
\(=2\sqrt{\frac{4-3}{2}}+4+2\sqrt{3}\)
\(=\sqrt{2}+4+2\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 6 2019
\(A=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\left(4-\sqrt{2}\right)}}}\)
\(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{4+2\sqrt{3}}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)^2=\sqrt{2}+4+2\sqrt{3}\)
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
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{4^2-2.4.\sqrt{2}+\sqrt{2^2}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\left|4-\sqrt{2}\right|}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left|\sqrt{3}-1\right|}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{3}-2}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\sqrt{3^2}-1^2\\ =3-1\\ =2\)