Question

[Sơn Thanh]

Viết các biểu thức sau dưới dạng bình phương của 1 tổng hay 1 hiệu

1/ 7-2√6

2/10 + 2√21

3/11 +2√33

4/10 + 4√16

5/11 - 2√30

6/11 + 2√10

7/11 - 4√6

8/ 11 + 4√7

9/12 - 4√5

10/ 12 + 6√3 

Answer 1

[huỳnh thị ngọc ngân]

1/ \(7-2\sqrt{6}=\left(\sqrt{6}\right)^2-2\sqrt{6}+1\)

\(=\left(\sqrt{6}-1\right)^2\)

2/ \(10+2\sqrt{21}=\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2\)

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

4/ \(10+4\sqrt{6}=2^2+2.2.\sqrt{6}+\left(\sqrt{6}\right)^2\)

\(=\left(2+\sqrt{6}\right)^2\)

5/ \(11-2\sqrt{30}=\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2\)

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

8/ \(11+4\sqrt{7}=2^2+2.2.\sqrt{7}+\left(\sqrt{7}\right)^2\)

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

10/ \(12+6\sqrt{3}=3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2\)

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