Question

Daca reusiti sa rezolvati vreun exercitiu scrieti numarul.

Answer 1

Varianta 2, exercitiul1.

Vom folosi definitia partii intregi:

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

Avem:

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

$3\leq3x<4$

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

Answer 2

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