Răspuns:
x₁ = 1 ; x₂ = 5
Explicație pas cu pas:
(3x-1)/(x+1) + (x+1)/(2x-1) = 3 <=>
(3x-1)(2x-1) + (x+1)(x+1) = 3(x+1)(2x-1) ; x ≠ { -1 ; 1/2}
6x²-3x-2x+1+x²+2x+1= 3(2x²-x+2x-1) <=>
7x²-3x+2 = 6x²-3x+6x-3 =>
7x²-6x²-3x-3x+2+3 = 0 =>
x²-6x+5 = 0 <=> x²-5x-x+5 = 0 <=>
x(x-5)-(x-5) = 0 <=> (x-5)(x-1) = 0 =>
x₁ = 1 ; x₂ = 5
