Question

[Anh Vi]

1.\(\sqrt{\frac{129}{16}+\sqrt{2}}\)

2.\(\sqrt{\frac{289+4\sqrt{72}}{16}}\)

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

4.\(\left(\sqrt{21}+7\right).\sqrt{10-2\sqrt{21}}\)

5.\(2.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{4+\sqrt{6-2\sqrt{5}}}\)

6.\(\left(4\sqrt{2}+\sqrt{30}\right).\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)

7.\(\left(7+\sqrt{14}\right).\sqrt{9-2\sqrt{14}}\)

Answer 1

[Nguyễn Phương Uyên]

1. Đặt A =\(\sqrt{\frac{129}{16}+\sqrt{2}}\)

\(\sqrt{16}\)A = \(\sqrt{129+16\sqrt{2}}\)

4A = \(\sqrt{\left(8\sqrt{2}+1\right)^2}\)

4A = \(8\sqrt{2}+1\)

⇒ A = \(\frac{\text{​​}8\sqrt{2}+1}{4}\)= \(2\sqrt{2}\) + \(\frac{1}{4}\)

2. Đặt B = \(\sqrt{\frac{289+4\sqrt{72}}{16}}\)

\(\sqrt{16}\)B = \(\sqrt{289+24\sqrt{2}}\)

4B = \(\sqrt{\left(12\sqrt{2}+1\right)^2}\)

4B = \(12\sqrt{2}+1\)

⇒ B = \(\frac{12\sqrt{2}+1}{4}\)= \(3\sqrt{2}+\frac{1}{4}\)

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

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

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

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

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

= \(\left(\sqrt{3}\right)^2\) - 1²

= 3 - 1

= 2

4. \(\left(\sqrt{21}+7\right)\). \(\sqrt{10-2\sqrt{21}}\)

= \(\left(\sqrt{21}+7\right)\) . \(\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

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

= \(\sqrt{7}\) \(\left[\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2\right]\)

= \(\sqrt{7}\) . (7 - 3)

= 4\(\sqrt{7}\)

5. \(2.\left(\sqrt{10}-\sqrt{2}\right)\). \(\sqrt{4+\sqrt{6-2\sqrt{5}}}\)

= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{4+\sqrt{5}-1}\)

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

= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{12+4\sqrt{5}}\)

= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\left(\sqrt{10}+\sqrt{2}\right)\)

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

= 10 - 2

= 8

6. \(\left(4\sqrt{2}+\sqrt{30}\right)\). \(\left(\sqrt{5}-\sqrt{3}\right)\). \(\sqrt{4-\sqrt{15}}\)

= \(\sqrt{2}\)\(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{4-\sqrt{15}}\)

= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{8-2\sqrt{15}}\)

= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\)

= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)^2\)

= \(\left(4+\sqrt{15}\right)\). \(\left(8-2\sqrt{15}\right)\)

= 32 - \(8\sqrt{15}\) + \(8\sqrt{15}\) - 30

= 2

7. \(\left(7-\sqrt{14}\right)\) . \(\sqrt{9-2\sqrt{14}}\)

= \(\sqrt{7}\) \(\left(\sqrt{7}-\sqrt{2}\right)\). \(\left(\sqrt{7}-\sqrt{2}\right)\)

= \(\sqrt{7}\). \(\left(\sqrt{7}-\sqrt{2}\right)^2\)

= \(\sqrt{7}\) . \(\left(9-2\sqrt{14}\right)\)

= 9\(\sqrt{7}\) - 14\(\sqrt{2}\)

TICK MÌNH NHA!

Answer 2

[Nguyễn Ngọc Khả Vy]

Bạn thông minh ghê!