Question

1)Comparati cu 0 numarul 2-$2 \sqrt{6}$
2)Se considera numerele M=$\sqrt{8}$ si N=$\frac{ \sqrt{2}+1 }{ \sqrt{2}-1 } }$ .
Aratati ca M<N

Answer 1

1)
2-2√6=2 - 2×2,...  = minus ceva. 
Deci 0 > 2-2√6

2)
$\frac{ \sqrt{2}+1 }{\sqrt{2}-1} = \frac{ (\sqrt{2}+1)^{2}}{ (\sqrt{2}-1)(\sqrt{2}+1)} = \frac{2-2 \sqrt{2}+1 }{2-1} = 3-2 \sqrt{2} \\ \sqrt{8} = 2 \sqrt{2} \\ \\ \Rightarrow 2 \sqrt{2} \square 3- 2 \sqrt{2}|-2 \sqrt{2} \Rightarrow 0\ \textless \ 3- 4 \sqrt{2} \\ \Rightarrow\sqrt{8} \ \textless \ \frac{ \sqrt{2}+1 }{\sqrt{2}-1}$