Răspunsuri

2014-06-16T23:23:36+03:00
Varianta 2, exercitiul1.

Vom folosi definitia partii intregi:

[x]=n\Leftrightarrow n\leq x<n+1,n\in\ N

Avem:

[3x+1]=4\Leftrightarrow 4\leq3x+1<5,\ x \in \ Z

3\leq3x<4

1\leq x<\dfrac43 ;\  x\in \ Z\Rightarrow x=1


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2014-06-17T01:09:19+03:00
Var. I 3
notam e^{x} =t
ecuatia devine
t+1/t=2
t²-2t+1=0
(t-1)²=0
t=1⇒ e^{x} = e^{0} ⇒x=0

varI 6
cosa=-2/3
sin²a+cos²a=1
sin²a=1-4/9=5/9⇒sin a=+-√5/3, deoarece a∈(π,3π/2)⇒sina =-√5/3
sin 2a=2sinacosa=4√5/9
cos 2a=cos²a-sin²a=-1/9
tg 2a=sin2a/cos2a=-4√5
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