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Comparați numerele reale
[tex]x = 4 \sqrt{2 {}^{26} + 2 {}^{27} } [/tex]
și
[tex]y = 8 \sqrt{2 {}^{26} + 2 {}^{24} } [/tex]
VĂ ROGGGG!!!!!!!DAU COROANĂĂĂĂĂ!!!!!!!!!​


Răspuns :

[tex]x = 4 \sqrt{ {2}^{26} + 2 {}^{27} } \\ x = \sqrt{4 {}^{2}(2 {}^{26} + 2 {}^{27}) } \\ x = \sqrt{(2 {}^{2}) {}^{2} ( {2}^{26} + 2 {}^{27}) } \\ x = \sqrt{2 {}^{4}(2 {}^{26} + 2 {}^{27} )} \\ x = \sqrt{2 {}^{4 + 26} + 2 {}^{4 + 27} } \\ \boxed{\huge{x = \sqrt{2 {}^{30} + 2 {}^{31} } }}[/tex]

[tex]y= 8\sqrt{ {2}^{26} + 2 {}^{24} } \\ y= \sqrt{8{}^{2}(2 {}^{26} + 2 {}^{24}) } \\ y= \sqrt{(2 {}^{3}) {}^{2} ( {2}^{26} + 2 {}^{24}) } \\ y = \sqrt{2 {}^{6}(2 {}^{26} + 2 {}^{24} )} \\ y = \sqrt{2 {}^{6 + 26} + 2 {}^{6+ 24} } \\ {y = \sqrt{2 {}^{32} + 2 {}^{30} } } \\ \huge \boxed{y = \sqrt{2 {}^{30} + 2 {}^{32} } }[/tex]

x<y

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