a/ \(11+6\sqrt{2}=9+2\cdot3\cdot\sqrt{2}+2=\left(3+\sqrt{2}\right)^2\left(đpcm\right)\)
b/ \(8-2\sqrt{7}=\sqrt{7}^2-2\sqrt{7}\cdot1+1=\left(\sqrt{7}-1\right)^2\left(đpcm\right)\)
Ta có:a)11+6\(\sqrt{2}\) =\(9+2.3.\sqrt{2}+2\) =\(3^2+2.3.\sqrt{2}+\left(\sqrt{2}\right)^2\)=\(\left(3+\sqrt{2}\right)^2\)
b) \(8-2\sqrt{7}\) = \(7-2.1.\sqrt{7}+1\) =\(\left(\sqrt{7}\right)^2-2.1.\sqrt{7}+1^2\) =\(\left(\sqrt{7}-1\right)^2\)