Răspuns:
a)
[tex]\displaystyle x \ = \frac{4}{\sqrt{2} } \ = \ \frac{4\sqrt{2} }{\sqrt{2} \ \cdot \ \sqrt{2}} \ = \ \frac{4\sqrt{2} }{2} \ = \ 2\sqrt{2} \\\\2\sqrt{2} \ = \ \sqrt{2^{2} \ \cdot \ 2 } \ = \sqrt{4 \ \cdot \ 2 } = \ \sqrt{8}\\ \\ x \ < \ y[/tex]
b)
[tex]\displaystyle x \ = - \ \frac{12}{ \sqrt{3} } \ = \ - \frac{12 \ \cdot \ \sqrt{3} }{ \sqrt{3} \ \cdot \ \sqrt{3} } \ = \ - \frac{12\sqrt{3} }{ 3} \ = \ - \ 4\sqrt{3} \ = \ - \ \sqrt{48} \\y = \ - \ \frac{10}{\sqrt{2} } \ = \ - \ \frac{10 \ \cdot \ \sqrt{2} }{\sqrt{2} \ \cdot \ \sqrt{2} } \ = \ - \ \frac{10\sqrt{2} }{2} \ = \ - \ 5\sqrt{2} \ = \ - \sqrt{50} \\\\x \ > \ y[/tex]
c)
[tex]\displaystyle x \ = \ \frac{7}{\sqrt{7} } \ + \ \sqrt{28} \ = \ \frac{7\sqrt{7} }{\sqrt{7} \ \cdot \ \sqrt{7}} \ + \ \sqrt{4 \ \cdot \ 7} \ = \ \frac{7\sqrt{7} }{7} \ + \ 2\sqrt{7} = \ \sqrt{7} \ + \ 2\sqrt{7}\ = \ 3\sqrt{7} \ = \sqrt{63} \\\\y \ = \ \frac{24\sqrt{3} }{ \sqrt{27} } \ = \ \frac{24\sqrt{3} }{ \sqrt{9 \ \cdot \ 3} } \ = \ \frac{24\sqrt{3} }{ 3 \sqrt{3} } = \ 8 \ = \ \sqrt{64} \\\\x \ < \ y[/tex]
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