conditii de ecistnta
2x-5>0 ,x>5/2 si 2x-1>0 x>1/2 ⇒x∈(5,2,+∞)
se transforma in
㏒(5) (2x-5)(2x-1)=㏒(5)5
(2x-5)(2x-1)=5
4x²-10x-2x+5=5
4x²-12x=0 |:4
x²-3x=0
x(x-3)=0
x=0 ∉domeniului
x=3 este solutie
lg(x+1)/(x-1)²=lg 10
ce: x+1>0 ⇒x>-1
(x-1)>0 ⇒x>1 si x-1≠0 ,x≠1
x∈(1,+∞)
(x+1)/(x-1)²=10
x+1=10(x-1)²
10x²-20x+10-x-1=0
10x²-21x+9=0
Δ=441-360=81
x1,2=(21+/-9)/20= 30/20=3/2
= 12/20∉ domeniului
