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Se considera functia f:R - > R, f(x) = x/2 +1

a) Verificati egalitatea f(rad 3) + f(3rad3)=2f(2rad3)

Se Considera Functia FR Gt R Fx X2 1a Verificati Egalitatea Frad 3 F3rad32f2rad3 class=

Răspuns :

Rayzen

[tex]f(x) = \dfrac{x}{2}+1\\ \\ f(\sqrt 3) +f(3\sqrt 3) = \Big(\dfrac{\sqrt 3}{2}+1\Big)+\Big(\dfrac{3\sqrt 3}{2}+1\Big) = \\ \\ = \dfrac{\sqrt 3+3\sqrt 3}{2}+2 = \dfrac{4\sqrt 3}{2}+2 = 2\sqrt 3+2 = 2\cdot \Big(\dfrac{2\sqrt 3}{2}+1\Big) = \\ \\ = 2f(2\sqrt 3)\quad (A)[/tex]

Răspuns:

Explicație pas cu pas:

2f(2[tex]\sqrt{3}[/tex])=2([tex]\frac{2\sqrt{3} }{2}[/tex]+1)

=2[tex]\sqrt{3}[/tex]+2

=2(1+[tex]\sqrt{3}[/tex])

f([tex]\sqrt{3}[/tex])+f(3[tex]\sqrt{3}[/tex])=[tex]\frac{\sqrt{3} }{2}[/tex]+1+[tex]\frac{3\sqrt{3} }{2}[/tex]+1

=[tex]\frac{4\sqrt{3} }{2}[/tex]+2

=2[tex]\sqrt{3}[/tex]+2

=2(1+[tex]\sqrt{3}[/tex])

Deci da...Se verifica egalitatea

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