Chứng minh đẳng thức : \(\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}=8\)
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TN
4
10 tháng 9 2019
\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{2}\cdot\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(3+\sqrt{5}\right)\cdot\left(\sqrt{5}-1\right)^2\)
\(=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
\(=2\cdot\left(3+\sqrt{5}\right)\cdot\left(3-\sqrt{5}\right)\)
\(=2\cdot\left(9-5\right)\)
\(=2-4=8\)
25 tháng 9 2019
\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right).\sqrt{2}.\sqrt{3-\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right).\sqrt{\left(\sqrt{5-1}\right)^2}\)
= \(\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)^2\)
= \(\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
= \(2.\left(3+\sqrt{5}\right).\left(3-\sqrt{5}\right)\)
= \(2.\left(9-5\right)\)
= \(2.4=8\)
Chúc bạn học tốt !!!
PT
19 tháng 7 2020
a. Sửa đề: \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
biến đổi vế trái :
ta có :\(\left(3+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3^2-\left(\sqrt{5}\right)^2}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
=2(\(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\))
=2(\(\sqrt{5}+5-\sqrt{5}-1\))
=2.4=8=VP
=> đpcm
b. Đặt vế trái là A
ta có \(A^2=\sqrt{2}+1-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\sqrt{2}-1\)
=\(2\sqrt{2}-2\)
=2\(\left(\sqrt{2}-1\right)\)
=> A=\(\sqrt{2\left(\sqrt{2}-1\right)}\)
vậy VT=VP =>đpcm
7 tháng 7 2021
Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{\sqrt{20}-2}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)}{2\left(\sqrt{5}-1\right)}-\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{6+2\sqrt{5}-\sqrt{5}}{2}\)
\(=\dfrac{6-\sqrt{5}}{2}\)
6 tháng 7 2021
Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{\sqrt{2}\left(\sqrt{6}+1\right)}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)
\(=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)\left(\sqrt{6}+1\right)-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)
\(=\dfrac{\sqrt{2}\left(3\sqrt{6}+3+\sqrt{30}+\sqrt{5}\right)-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)
\(=\dfrac{6\sqrt{3}+3\sqrt{2}+2\sqrt{15}+\sqrt{10}-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)
\(=\dfrac{6\sqrt{3}+3\sqrt{2}+\sqrt{15}+\sqrt{10}-\sqrt{5}}{ }\)
Đề sai rồi bạn
NH
19 tháng 8 2019
a, \(VT=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{20}-2\right)}{2}\)
\(=\frac{\sqrt{5-2\sqrt{5}+1}\left(3+\sqrt{5}\right)\left(2\sqrt{5}-2\right)}{2}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)2\left(\sqrt{5}-1\right)}{2}\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18-6\sqrt{5}+6\sqrt{5}-10=8=VP\)
b, \(VT=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left(16-15\right)=2=VP\)
Biến đổi vế trái
\(\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)=\(\left(\sqrt{3+\sqrt{5}}\right)^2.\sqrt{3-\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
=\(\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{4}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{10\left(3+\sqrt{5}\right)}-2\sqrt{2\left(3+\sqrt{5}\right)}\)
\(=2\sqrt{30+10\sqrt{5}}-2\sqrt{6+2\sqrt{5}}\)
\(=2\sqrt{\left(5+\sqrt{5}\right)^2}-2\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=2\left(5+\sqrt{5}\right)-2\left(\sqrt{5}+1\right)\)
\(=10+2\sqrt{5}-2\sqrt{5}-2=8\)
Sau khi biến đổi ta thấy vế trái bằng vế phải. Vậy đẳng thức đã được chứng minh