ex. 22
E( x) = [ ( _x+ 3)²_ + 1 + _2x+ 6_] ·_ x- 3_ - 1
(x-3)² ( x- 3) 4x
a. x= ? , E ≠ sens
x- 3= 0 4·x= 0
x = 3 x= 0
b. E( x) = [ ( _x+ 3)²_ + 1 + _2x+ 6_] ·_ x- 3_ - 1
(x-3)² ( x- 3) 4x
x- 3≠ 0, x ≠ 3
4·x≠ 0 , x ≠ 0
E( x) = [ ( _x+ 3)²_ + 1 (x-3)²+ _(2x+ 6)(x-3)_] ·_ x- 3_ - 1
(x-3)² 4x
E( x) = [ _x²+6x+9+ x²-6x+ 9+2x²-18_] ·_ x- 3_ - 1
(x-3)² 4x
E( x) = [ _x²+9+ x²+ 9+2x²-18_] ·_ x- 3_ - 1
(x-3)² 4x
E( x) = [ _x²+ x²+2x²] ·_ x- 3_ - 1
(x-3)² 4x
E( x) = [ _4x²] ·_ x- 3_ - 1
(x-3)² 4x
E( x) = _x_ - 1
x-3
E (x) = _x- (x-3)_
x-3
E(x) =_x- x+ 3_
x-3
E ( x) = _3_
x-3
c. a∈ Z , E ( a) ∈ Z
E ( a) = _3_
a- 3
S₁: a- 3= 1 S₂: a- 3= -1 S₃: a- 3= 3 S₄: a- 3= -3
a = 4 a =2 a = 6 a =0
d. a∈ Z , E ( a) ∈ Z : 3
S₁: a- 3= 1 S₂: a- 3= -1
a = 4 a =2
a)
[tex]\it E(x)\ nu\ are\ sens\ pentru\ valorile\ lui\ x\ care\ anuleaz\breve{a}\ numitorii.\\ \\ x-3=0 \Rightarrow x=3\\ \\ 4x=0 \Rightarrow x=0\\ \\ Deci,\ E(x)\ nu\ are\ sens\ pentru\ x\in\{0,\ 3\}[/tex]
b)
[tex]\it Not\breve{a}m\ \ \dfrac{x+3}{x-3}=t,\ iar\ expresia\ din\ paranteza\ dreapt\breve{a}\ devine:\\ \\ \\ t^2+1+2t=t^2+2t+1=(t+1)^2\\ \\ \\ Revenim\ asupra\ nota\c{\it t}iei,\ iar\ expresia\ din\ paranteza\ dreapt\breve{a}\ devine:\\ \\ \Big(\dfrac{x+3}{x-3}+1\Big)^2=\Big(\dfrac{x+3+x-3}{x-3}\Big)^2=\dfrac{4x^2}{(x-3)^2}[/tex]
Acum, expresia din enunț se scrie:
[tex]\it E(x)=\dfrac{4x^2}{(x-3)^2}\cdot\dfrac{x-3}{4x}-1=\dfrac{x}{x-3}-1=\dfrac{x-x+3}{x-3}=\dfrac{3}{x-3}[/tex]
c)
[tex]\it E(a)=\dfrac{3}{a-3}\in\mathbb{Z} \Rightarrow a-3\in D_3 \Rightarrow a-3\in\{-3,\ -1,\ 1,\ 3\}|_{+3}\Rightarrow \\ \\ \Rightarrow a\in\{0,\ 2,\ 4,\ 6\}[/tex]
d)
Calculăm E(a), pentru a ∈ {0, 2, 4, 6} și vom determina :
E(2) = -3 ∈ ℤ, E(4) = 3 ∈ ℤ
Prin urmare, pentru a ∈ {2, 4} ⇒ E(a) ∈ {-3, 3} ⇒ 3 | E(a)
