Thực hiện phép tính:\(\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{12}\right)\)
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5 tháng 8 2021
a) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\dfrac{2-\sqrt{3}}{1}+\dfrac{2+\sqrt{3}}{1}\)
=4
5 tháng 8 2021
\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(=\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(5-2\sqrt{6}\right)\)
\(=\sqrt{3}-\sqrt{2}\)
LH
1
25 tháng 6 2018
Đây phải là dấu cộng
B=\(\sqrt{3+\sqrt{5}}.\left(3+\sqrt{5}\right).\sqrt{2}.\left(\sqrt{5}-1\right)\)
B=\(\sqrt{6+2\sqrt{5}}.\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)\)
B=\(\sqrt{\left(\sqrt{5}+1\right)^2}.\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)\)
B=\(\left(\sqrt{5}+1\right).\left(\sqrt{5}-1\right).\left(3+\sqrt{5}\right)\)
B=12+4\(\sqrt{5}\)
PA
7 tháng 8 2017
\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)^2\)
\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)
\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)
https://hoc24.vn/hoi-dap/question/405366.html
\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
= 5 - 3
= 2
2 tháng 2 2022
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
\(\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}\times\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{6+2\sqrt{5}}\times\left(\sqrt{5}-1\right)\times\sqrt{9-5}\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{5}-1\right)\)
\(=2\left(5-1\right)\)
= 8