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
22 tháng 6 2018
\(B=\left(\sqrt{10}+\sqrt{6}\right).\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\) (vì\(\sqrt{5}-\sqrt{3}>0\))
\(=2\sqrt{2}\)
22 tháng 6 2018
\(A=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right).\sqrt{2}\)
\(=\sqrt{4}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{4}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{4}-\left|\sqrt{5}-1\right|\)
\(=\sqrt{4}-\sqrt{5}+1\) (vì \(\sqrt{5}-1>0\))
NM
1
8 tháng 9 2019
\(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right):\sqrt{5}\)
\(=\sqrt{20}:\sqrt{5}-\sqrt{45}:\sqrt{5}+\sqrt{5}:\sqrt{5}\)
\(=2-3+1\)
\(=0\)
KN
25 tháng 6 2019
b)\(\frac{\sqrt{27}}{\sqrt{12}}+\frac{1}{2}\)
\(=\frac{\sqrt{3}.\sqrt{9}}{\sqrt{3}.\sqrt{4}}+\frac{1}{2}\)
\(=\frac{\sqrt{9}}{\sqrt{4}}+\frac{1}{2}\)
\(=\frac{3}{2}+\frac{1}{2}\)
\(\frac{4}{2}=2\)
KN
25 tháng 6 2019
a) \(\sqrt{45}.\sqrt{15}.\sqrt{27}\)
\(=\left(\sqrt{15}\right)^2.\left(\sqrt{3}\right)^2.\sqrt{9}\)
\(=15.3.3\)
\(=135\)
23 tháng 7 2019
\(a,\sqrt{\frac{72}{9}}:\sqrt{8}=\frac{\sqrt{72}}{\sqrt{9}}.\frac{1}{\sqrt{8}}\)
\(=\frac{6\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}\)
\(=1\)
\(b,\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
\(=33\sqrt{3}:\sqrt{3}\)
\(=33\)
\(c,\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\left(5\sqrt{5}+7\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)
\(=11\sqrt{5}:\sqrt{5}\)
\(=11\)
\(d,\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{7}{\sqrt{7}}\right):\sqrt{7}\)
\(=\frac{4}{\sqrt{7}}.\frac{1}{\sqrt{7}}=\frac{4}{7}\)
4 tháng 8 2016
a) \(\sqrt{21-6\sqrt{6}}-\sqrt{9+2\sqrt{18}}\)
\(=\sqrt{18-2\sqrt{18\cdot3}+3}-\sqrt{6+2\sqrt{18}+3}\)
\(=\left(\sqrt{18}-\sqrt{3}\right)^2-\left(\sqrt{6}-\sqrt{3}\right)^2\)
\(=\sqrt{18}-\sqrt{3}-\sqrt{6}+\sqrt{3}\)
\(=\sqrt{18}+\sqrt{6}=\sqrt{6}\left(\sqrt{3}+1\right)\)
12 tháng 8 2020
\(A=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{4+2\sqrt{3}}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
=> \(A=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)
=> \(A=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
=> \(A=4\sqrt{3}-8+6-4\sqrt{3}\)
=> \(A=-8+6=-2\)
VẬY \(A=-2\)
12 tháng 8 2020
\(B=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2}.\sqrt{4-\sqrt{15}}\)
=> \(B=\sqrt{8-2\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
=> \(B=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\left(4+\sqrt{15}\right)\)
=> \(B=\left(\sqrt{5}-\sqrt{3}\right)^2\left(4+\sqrt{15}\right)\)
=> \(B=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
=> \(B=32+8\sqrt{15}-8\sqrt{15}-30\)
=> \(B=2\)
VẬY \(B=2\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
24 tháng 8 2016
\(=\left(\sqrt{2.3}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}.\)
\(=\left(3\sqrt{2}-2\sqrt{2.3}+\sqrt{2.3}-2\sqrt{2}\right)\sqrt{\sqrt{3}+2}\)
\(=\left(\sqrt{2}-\sqrt{2.3}\right)\sqrt{\sqrt{3}+2}=\sqrt{2}\left(1-\sqrt{3}\right)\sqrt{\sqrt{3}+2}\)
\(\sqrt{0,45.0,3.6}\)
\(=\sqrt{0,135.6}\)
\(=\sqrt{0,81}\)
\(=0,9\)