Question

Rezolvati ecuatiile reductibile la gradul 2:
1/x+1 + 2/x+2=2

Answer 1

Numitor comun (x+1)(x+2)=x²+3x+2
[(x+2)/x²+3x+2] + [(2x+2)/x²+3x+2] = [2(x²+3x+2)]/x²+3x+2   eliminam numitorul
x+2 + 2x+2 = 2(x²+3x+2)
3x+4 = 2x²+6x+4
2x²+6x+4-3x-4=0
2x²+3x=0
x(2x+3)=0
x₁=0
2x+3=0    ⇒   2x= - 3     ⇒     x₂= -3/2

Answer 2

banuiesc ca ai
1/(x+1) +2/(x+2)=2

fie x+1=y
attunci
1/y+  2/y(y+1)=2


(y+1+2y)/y(y+1)=2
3y+1=2y²+2y
2y²-y-1=0
y1,2=(1+-√9)/4=(1+-3)/4
y1=-2/4=-1/2
y2=4/4=1

x+1=-1/2...x=-3/2  ( verificare -2+4=2)
x+1=1.......x=0 (caz care duce  la verificarea 1+1=2.....:::))  )