Tính:
a) \(A=\left(\sqrt{6}+\sqrt{10}\right)-\sqrt{4-\sqrt{15}}\)
b) \(B=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{15}}\)
c) \(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
K
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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
MT
10 tháng 7 2015
\(A=\left(\sqrt{5}-\sqrt{2}\right)^2+2\sqrt{10}=\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^2+2\sqrt{10}\)
=\(5-2\sqrt{10}+2+2\sqrt{10}=7\)
câu b hình như sai đề
\(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)=\sqrt{3-\sqrt{5}}\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{2\left(3-\sqrt{5}\right)}\left(\sqrt{5}-1\right)=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\)
=\(\sqrt{5-2\sqrt{5}+1}\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\)
=\(\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)=\left(\sqrt{5}-1\right)^2=\left(\sqrt{5}\right)^2-2\sqrt{5}+1\)
=\(6-2\sqrt{5}\)
10 tháng 7 2015
\(A=\left(\sqrt{5}-\sqrt{2}\right)^2+2\sqrt{10}=\left(\sqrt{5.1}-\sqrt{2.1}\right)^2+2\sqrt{10}=7-2\sqrt{10}+2\sqrt{10}=7\)
Câu B hình như có cái gì đó không ổn
\(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)=\sqrt{3-\sqrt{5}}\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2\left(3-\sqrt{5}\right)}\left(\sqrt{5-1}\right)=\sqrt{6-2\sqrt{5}}\left(-1+\sqrt{5}\right)=6-2\sqrt{5}\)
BL
12 tháng 11 2017
\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
BL
12 tháng 11 2017
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 8 2020
a/ \(=\sqrt{\left(\sqrt{3}-1\right)^2\left(2\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)=5-\sqrt{3}\)
b/ \(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}+1\right)=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)^2\)
\(=\left(\sqrt{3}-2\right)\left(4+2\sqrt{3}\right)=2\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\)
\(=2\left(3-4\right)=-2\)
c/ \(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
\(=2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=2.\left(9-5\right)=8\)
d/ \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)
Tính
a) Ta có: \(A=\left(\sqrt{6}+\sqrt{10}\right)-\sqrt{4-\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{4-\sqrt{15}}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\sqrt{5}-\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{3}+\sqrt{5}-\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{5}+\sqrt{3}\)
\(=2\sqrt{3}\)
c) Ta có: \(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\cdot\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left[4^2-\left(\sqrt{15}\right)^2\right]\)
\(=2\cdot\left[16-15\right]=2\cdot1=2\)