Question

Đưa về hằng đẳng thức:

a, \(4-2\sqrt{3}\)

b, \(3+2\sqrt{2}\)

c, \(11-2\sqrt{30}\)

Answer 1

a, 4-2\(\sqrt{3}\)

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

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

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

b,3+\(2\sqrt{2}\)

=\(2+2\sqrt{2}.\sqrt{1}+1\)

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

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

c, 11-2\(\sqrt{30}\)

=6-\(2\sqrt{5}.\sqrt{6}+5\)

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

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

Answer 2

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

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

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