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Tuấn Anh Nguyễn
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Nguyễn Như Nguyệt
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Nguyễn Lê Phước Thịnh
19 tháng 5 2022 lúc 14:00

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)

\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)

hương giang
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Phong Kinn
3 tháng 9 2019 lúc 20:34

a) \(\sqrt{3^2}-\sqrt{7^2}+\sqrt{\left(-1\right)^2}=|3|-|7|+|-1|=3-7+1=-3\)

b) \(-2\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}+\sqrt{3^2}=-2|2|+|-5|+\left|3\right|=-4+5+3=4\)

c) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2-\sqrt{2}\right|+\left|2+\sqrt{2}\right|=2-\sqrt{2}+2+\sqrt{2}=4\)

d) \(\sqrt{\left(3\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}=\left|3\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3\sqrt{2}-\sqrt{2}+1=2\sqrt{2}+1\)

e) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}-1\right|+\left|\sqrt{2}+1\right|=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)

f) \(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|+\left|\sqrt{5}+2\right|=\sqrt{5}-2+\sqrt{5}+2=2\sqrt{5}\)

g) \(\sqrt{9-4\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{9-2\sqrt{8}}+\sqrt{2-2\sqrt{2}.3+9}=\sqrt{\left(\sqrt{8}-1\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}=\sqrt{8}-1+3-\sqrt{2}=2-\sqrt{2}+\sqrt{8}\)

h) \(\sqrt{12+8\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{12+2\sqrt{4}\sqrt{8}}+\sqrt{6-2\sqrt{2}\sqrt{4}}=\sqrt{\left(\sqrt{4}+\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}=\sqrt{4}+\sqrt{8}+\sqrt{4}-\sqrt{2}\)

k) \(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{\left(\sqrt{3}+2\right)^2}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)

Anh Quynh
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Tuấn Tú
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YangSu
26 tháng 6 2023 lúc 13:57

Yêu cầu?

Hotboy nguyên
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Nguyễn Lê Phước Thịnh
19 tháng 5 2022 lúc 23:34

b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)

c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)

d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)

Nguyễn Hoàng Anh Thư
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nguyen minh huyen
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💋Bevis💋
17 tháng 7 2019 lúc 10:02

\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)

Ta có

:\(VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)

\(=|2-\sqrt{5}|-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}\)

\(=-2=VP\left(đpcm\right)\)

\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)

Ta có:

\(VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)

\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

\(=\frac{2+\sqrt{2}+\sqrt{2}+1}{\sqrt{2}^2-1^2}\)

\(=\frac{3+2\sqrt{2}}{2-1}\)

\(=3+2\sqrt{2}=VP\left(đpcm\right)\)

💋Bevis💋
17 tháng 7 2019 lúc 10:19

c,Bạn xem lại đề

\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)

Ta có:

\(VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)

\(=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}\)

\(=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}\)

\(=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}\)

\(=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)

\(=\frac{8}{5-4}\)

\(=8=VP\left(đpcm\right)\)

💋Bevis💋
17 tháng 7 2019 lúc 10:32

\(e,\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)

\(VT=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3-\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)|1-\sqrt{5}|\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2\)

\(=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)

\(=\left(3+\sqrt{5}\right).2\left(3-\sqrt{5}\right)\)

\(=[3^2-\left(\sqrt{5}\right)^2].2\)

\(=4.2=8=VP\left(đpcm\right)\)

nguyen minh huyen
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