Question

viết thành bình phương của 1 biểu thức:

a,\(3\pm2\sqrt{2}\)

b,\(6\pm2\sqrt{5}\)

c,\(7\pm4\sqrt{3}\)

Answer 1

\(a,\) 

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

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

\(b,\)

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

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

\(c,\)

\(7+4\sqrt{3}=4+2.2\sqrt{3}+3=2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(2+\sqrt{3}\right)^2\)

\(7-4\sqrt{3}=2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(2-\sqrt{3}\right)^2\)

Answer 2

`a)3+-2sqrt2`

`=2+-2sqrt2+1`

`=(sqrt2+-1)^2`

`b)6+-2sqrt5`

`=5+-2sqrt5+1`

`=(sqrt5+-1)^2`

`7)7+-4sqrt3`

`=4+-2.2.sqrt3+3`

`=(2+-sqrt3)^2`