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a: \(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)

\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)

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

b: \(=\sqrt{2+\sqrt{2}}\cdot\sqrt{3+\sqrt{7+\sqrt{2}}}\cdot\sqrt{9-6-\sqrt{7+\sqrt{2}}}\)

\(=\sqrt{2+\sqrt{2}}\cdot\sqrt{9-7-\sqrt{2}}\)

\(=\sqrt{2}\)

20 tháng 9 2021

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

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

c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)

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

a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)

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

=-4

b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)

\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)

\(=3\sqrt{3}+1\)

c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)

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

\(=3\sqrt{5}-6\)

d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)

\(=\sqrt{7}-2+4-\sqrt{7}+8\)

=10

25 tháng 10 2023

a: \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)

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

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

b: Sửa đề: \(\dfrac{7-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)

\(=\dfrac{\sqrt{7}\left(\sqrt{7}-2\right)}{-\left(\sqrt{7}-2\right)}+\dfrac{6\left(\sqrt{7}-1\right)}{6}+18-12\)

\(=-\sqrt{7}+\sqrt{7}-1+6=5\)

AH
Akai Haruma
Giáo viên
11 tháng 8 2021

1.

\(\frac{3\sqrt{5}-5\sqrt{3}}{\sqrt{15}-3}=\frac{3\sqrt{5}-\sqrt{5}.\sqrt{15}}{\sqrt{15}-3}=\frac{-\sqrt{5}(\sqrt{15}-3)}{\sqrt{15}-3}=-\sqrt{5}\)

2.

\(\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2+2\sqrt{2.3}+3}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{(\sqrt{2}+\sqrt{3})^2}}{\sqrt{2}+\sqrt{3}}\)

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

3.

\(\frac{7+4\sqrt{3}}{2+\sqrt{3}}=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)

AH
Akai Haruma
Giáo viên
11 tháng 8 2021

4.

\(\frac{16-6\sqrt{7}}{\sqrt{7}-3}=\frac{3^2-2.3\sqrt{7}+7}{\sqrt{7}-3}=\frac{(\sqrt{7}-3)^2}{\sqrt{7}-3}=\sqrt{7}-3\)

5.

\(\frac{(\sqrt{3}-\sqrt{2})^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{3+2+2\sqrt{2.3}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2})^2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}\)

6.

\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{6-2\sqrt{10}}}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{6-2\sqrt{10}}}\)

NV
19 tháng 8 2021

\(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

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

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

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

\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)

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

1: Ta có: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

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

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

2: Ta có: \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

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

\(=2\sqrt{3}\)