Explicație pas cu pas:
[tex]\boxed{ \red{ {(a \pm b)}^{2} = {a}^{2} \pm 2ab + {b}^{2} }} [/tex]
[tex]\boxed{ \red{ (a - b)(a + b) = {a}^{2} - {b}^{2} }} [/tex]
a)
[tex]{\Big( \sqrt{4 + \sqrt{5} } + \sqrt{4 - \sqrt{5} } \Big)}^{2} = 4 + \sqrt{5} + 2 \sqrt{(4 + \sqrt{5})(4 - \sqrt{5})} + 4 - \sqrt{5} = 8 + 2 \sqrt{16 - 5} = \bf 8 \sqrt{11} [/tex]
b)
[tex]{\Big( \sqrt{\sqrt{2} + 1 } - \sqrt{\sqrt{2} - 1} \Big)}^{2} = \sqrt{2} + 1 - 2 \sqrt{(\sqrt{2} + 1)(\sqrt{2} - 1)} + \sqrt{2} - 1 = 2 \sqrt{2} - 2\sqrt{2 - 1} = 2 \sqrt{2} - 2\sqrt{1} = \bf 2 \sqrt{2} - 2 = 2( \sqrt{2} - 1) [/tex]
c)
[tex]{\Big( \sqrt{3 + 2\sqrt{2} } + \sqrt{3 - 2\sqrt{2} } \Big)}^{2} = 3 + 2 \sqrt{2} + 2 \sqrt{(\sqrt{3 + \sqrt{8} })(\sqrt{3 - \sqrt{8} })} + 3 - 2 \sqrt{2} = 6 + 2 \sqrt{9 - 8} = 6 + 2 \sqrt{1} = 6 + 2 = \bf 8[/tex]
d)
[tex]{\Big( \sqrt{2\sqrt{3} + 3} - \sqrt{2\sqrt{3} - 3} \Big)}^{2} = 2 \sqrt{3} + 3 - 2 \sqrt{(\sqrt{12} + 3)(\sqrt{12} - 3)} + 2 \sqrt{3} - 3 = 4 \sqrt{3} - 2 \sqrt{12 - 9} = 4 \sqrt{3} - 2 \sqrt{3} = \bf 2 \sqrt{3} [/tex]
