Question

Rezolvati ecuatiile:
(x+2)la a doua=5(x+2) x apartine lui R
x la a doua -3x=0 X apartine lu Q
5 |X| = |X|+1 X apartine lui R
2x la a doua -1=3X la a doua +1 X partine LUI R

Answer 1

x^2-3x=0
x(x-3)=0
x=0
x=3

x^2+4x+4=5x+10
x^2+4x-5x+4-10=0
x^2-x-6=0
a=1
b=2
c=-6
D=b^2-4ac
D=1+24
D=25
D=5
faci x1,x2

Answer 2

(x+2)^2=5(x+2)
Folosind formula (a+b)^2= a^2+2ab+b^2 rezulta ca:
(x+2)^2=x^2+2*x*2+2^2= x^2+4x+4
=> x^2+4x+4=5x+10
=> x^2+4x+4-5x-10=0
=> x^2-x-6=0
Folosim delta
d = b^2- 4*a*c
d = (-1)^2- 4*1*(-6)= 1+24=25
x1,2 = -b - sau + radical din delta /2a
x1= -4/2=2
x2= 6/2=3

S: x E {2,3} E(apartine)
x^2-3x=0
dam pe x factor comun
=>x(x-3)=0
x e 0 cand produsul este 0
x1=0
sau
x2=3

2x^2-1=3x^2+1
=>2x^2-1-3x-1=0
=>2x^2-3x-2=0
=> d = (-3)^2 - 4 * 2*(-2)= 9+16=25
x1,2 =
x1=3+5/2=8/2=4
x2=3-5/2=-2/2=-1


S: x E {-1,2}