Rut gon:
\(B=\frac{\sqrt{8+\sqrt{40+8\sqrt{5}}}+\sqrt{8-\sqrt{40+8\sqrt{5}}}}{\sqrt{50}+\sqrt{250}}\)
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14 tháng 6 2018
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
18 tháng 7 2017
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{8+\sqrt{2.4}+\sqrt{5.4}+\sqrt{10.4}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{1}\right)^2+\left(\sqrt{2}\right)^2+\left(\sqrt{5}\right)^2+2.\sqrt{2}.\sqrt{1}+2\sqrt{1}.\sqrt{5}+2\sqrt{5}.\sqrt{2}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{1}+\sqrt{2}+\sqrt{5}\right)^2}\)
= \(\sqrt{1}+\sqrt{2}+\sqrt{5}\)
18 tháng 7 2017
phần trên mk làm thiếu \(-\sqrt{2}-\sqrt{5}\)
kết quả là 1 mới đúng
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
7.
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+2\sqrt{4.3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{(\sqrt{4}+\sqrt{3})^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(2+\sqrt{3})}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-2.5\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5(5-\sqrt{3})}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
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.
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)
\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{(\sqrt{2}+\sqrt{5})^2}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}-\sqrt{(\sqrt{2}+\sqrt{5})^2}=|\sqrt{2}+\sqrt{5}+1|-|\sqrt{2}+\sqrt{5}|=1\)
2 tháng 7 2019
\(a,\left(\sqrt{27}-2\sqrt{17}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=3\sqrt{21}-2\sqrt{119}+7+7\sqrt{8}\)
Đề sai chăng???
2 tháng 7 2019
\(b,\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{2+2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{2}-1+\sqrt{2}+1\)
\(=2\sqrt{2}\)
\(c,9\sqrt{2}-4\sqrt{8}-\sqrt{50}+2\sqrt{32}\)
\(=9\sqrt{2}-8\sqrt{2}-5\sqrt{2}+8\sqrt{2}\)
\(=\sqrt{2}\left(9-8-5+8\right)\)
\(=4\sqrt{2}\)
\(d,\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{4+2.2\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=\sqrt{2}-1-2-\sqrt{2}\)
\(=-3\)
Mẫu số bằng \(\sqrt{50}+\sqrt{250}=5\sqrt{2}+5\sqrt{10}=5\sqrt{2}\left(1+\sqrt{5}\right).\)
Kí hiệu tử số là \(A\) thì ta có
\(A^2=\left(\sqrt{8+\sqrt{40+8\sqrt{5}}}+\sqrt{8-\sqrt{40+8\sqrt{5}}}\right)^2\)
\(=8+\sqrt{40+8\sqrt{5}}+2\sqrt{8+\sqrt{40+8\sqrt{5}}}\cdot\sqrt{8-\sqrt{40+8\sqrt{5}}}+8-\sqrt{40+8\sqrt{5}}\)
\(=16+2\sqrt{\left(8+\sqrt{40+8\sqrt{5}}\right)\left(8-\sqrt{40+8\sqrt{5}}\right)}\)
\(=16+2\sqrt{8^2-\left(40+8\sqrt{5}\right)}=16+2\sqrt{24-8\sqrt{5}}\)
\(=16+2\sqrt{4-2\cdot2\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}=16+2\sqrt{\left(2-2\sqrt{5}\right)^2}\)
\(=16+2\left|2-2\sqrt{5}\right|=16-4+4\sqrt{5}=12+4\sqrt{5}=4\left(3+\sqrt{5}\right).\)
Vậy \(A=4\left(3+\sqrt{5}\right)=2\left(6+2\sqrt{5}\right)=2\left(\sqrt{5}+1\right)^2.\)
Thành thử \(B=\frac{2\left(\sqrt{5}+1\right)^2}{5\sqrt{2}\left(1+\sqrt{5}\right)}=\frac{\sqrt{2}\left(\sqrt{5}+1\right)}{5}=\frac{\sqrt{10}+\sqrt{2}}{5}.\)