[tex] \it 5) \ \ m_a\geq m_g \Rightarrow \dfrac{a+b}{2} \geq\sqrt{ab} \Rightarrow a+b\geq2\sqrt{ab}
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Pentru\ b=\dfrac{1}{a},\ inegalitatea\ devine \ a+\dfrac{1}{a} \geq2\sqrt{a\cdot\dfrac{1}{a}} \Rightarrow a+\dfrac{1}{a} \geq2 \ (A) [/tex]
4)
[tex] \it A = \dfrac{^{\sqrt6-\sqrt5)}2\sqrt2}{\ \ \sqrt6+\sqrt5} =\dfrac{2\sqrt{12} -2\sqrt10}{(\sqrt6)^2-(\sqrt5)^2} = \dfrac{2\sqrt{4\cdot3}-2\sqrt{10}}{6-5} = 4\sqrt3-2\sqrt{10}
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A+2\sqrt{10} = 4\sqrt3-2\sqrt{10}+2\sqrt{10} = 4\sqrt3 [/tex]
3)
[tex] \it 8-2\sqrt{15} = 5+3-2\sqrt{15} =5-2\sqrt{15}+3 =(\sqrt5)^2-2\sqrt{15}+(\sqrt3)^2=
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= (\sqrt5-\sqrt3)^2
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Analog,\ \ 8+2\sqrt{15} = (\sqrt5+\sqrt3)^2 [/tex]
[tex] \it N = \sqrt{(\sqrt5-\sqrt3)^2} +\sqrt{(\sqrt5-\sqrt3)^2} =\sqrt5-\sqrt3+\sqrt5+\sqrt3 = 2\sqrt5
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(N-2\sqrt5)^{2014} =(2\sqrt5-2\sqrt5)^{2014} =0^{2014} =0 [/tex]
