NM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
BT
2 tháng 8 2017
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
29 tháng 8 2019
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
29 tháng 8 2022
a: \(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right)\)
\(=0\)
b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{8}-2\sqrt{5}-6\sqrt{5}\right)\)
\(=\sqrt{\sqrt{3}}\left(4\sqrt{2}-8\sqrt{5}\right)\)
1 tháng 2 2022
a: \(=2\sqrt{\sqrt{3}}\cdot4\sqrt{5}-2\cdot\sqrt{\sqrt{3}}\cdot\sqrt{5}-3\cdot\sqrt{\sqrt{3}}\cdot2\sqrt{5}\)
\(=2\sqrt{\sqrt{3}}\left(4\sqrt{5}-\sqrt{5}-3\sqrt{5}\right)=0\)
b: \(=2\cdot2\sqrt{2}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(2\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)=2\sqrt{\sqrt{3}}\cdot\left(2\sqrt{2}-4\sqrt{5}\right)\)
PK
18 tháng 8 2016
a/ \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}=2\sqrt{4.2.5\sqrt{4.3}}-2\sqrt{\sqrt{25.3}}-3\sqrt{5\sqrt{16.3}}\)
= \(2.2\sqrt{2.5.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}=4.2\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3.2\sqrt{5\sqrt{3}}\)
= \(\sqrt{5\sqrt{3}}\left(8-2-6\right)=\sqrt{5\sqrt{3}}.0=0\)
PK
18 tháng 8 2016
b/ \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}=2\sqrt{2.4\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{4.5\sqrt{3}}\)
= \(4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
15 tháng 7 2018
\(a.\sqrt{8-\sqrt{28}}+\sqrt{21+12\sqrt{3}}=\sqrt{7-2\sqrt{7}+1}+\sqrt{12+2.2\sqrt{3}.3+9}=\sqrt{7}-1+2\sqrt{3}+3=2\sqrt{3}+\sqrt{7}+2\) \(b.\sqrt{5+\sqrt{24}}-\sqrt{57-40\sqrt{2}}=\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}-\sqrt{32-2.4\sqrt{2}.5+25}=\sqrt{3}+\sqrt{2}-4\sqrt{2}+5=\sqrt{3}-3\sqrt{2}+5\) \(c.\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=4\sqrt{2}+2\sqrt{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 3 2019
a/ \(2\sqrt{10}-10\sqrt{10}+9\sqrt{10}=\sqrt{10}\)
b/ \(\frac{-1\left(4-3\sqrt{2}\right)+1\left(4+3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\frac{-4+3\sqrt{2}+4+3\sqrt{2}}{16-18}=\frac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c/ \(\left(3+\sqrt{5}\right).\sqrt{2}.\sqrt{7-3\sqrt{5}}=\left(3+\sqrt{5}\right)\sqrt{14-6\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\sqrt{\left(3-\sqrt{5}\right)^2}=\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=9-5=4\)
d/ \(3\sqrt{2}-4\sqrt{2}+5\sqrt{2}=4\sqrt{2}\)
e/ \(\sqrt{19+8\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(4+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=4+\sqrt{3}+2-\sqrt{3}=6\)
12 tháng 7 2018
\(1a.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{548}=2\sqrt{16.5\sqrt{3}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=8\sqrt{\sqrt{75}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=6\sqrt{\sqrt{75}}-6\sqrt{137}\) \(b.\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right).\dfrac{1}{\sqrt{15}}=10\sqrt{3}.\dfrac{1}{\sqrt{3}.\sqrt{5}}=2\sqrt{5}\) \(d.\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}=\left(60\sqrt{2}-80\sqrt{2}+105\sqrt{2}\right).\dfrac{1}{\sqrt{10}}=85\sqrt{2}.\dfrac{1}{\sqrt{2}.\sqrt{5}}=17\sqrt{5}\) \(e.\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right):\sqrt{7}=\left(\sqrt{\dfrac{1}{7}}-4\sqrt{\dfrac{1}{7}}+3\sqrt{\dfrac{1}{7}}\right).\dfrac{1}{\sqrt{7}}=0\) \(2a.A=\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}=\sqrt{9-5-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\) \(b.B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{4-2}=2\)
20 tháng 7 2017
c) \(\left(\sqrt{10}+\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)\left(6-2\sqrt{5}\right)\)
= \(\left(\sqrt{5}+1\right)\sqrt{2}\left(\sqrt{3+\sqrt{5}}\right)\left(5-2\sqrt{5}+1\right)\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{6+2\sqrt{5}}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{5+2\sqrt{5}+1}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{\left(\sqrt{5}+1\right)^2}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2\)
= \(4.4=16\)
d) \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{1+2+5+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{2}+\sqrt{1}+\sqrt{5}\right)^2}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{2}+\sqrt{5}+1-\sqrt{2}-\sqrt{5}\)
= \(1\)
\(\approx5,113080631\)