Log5(x+1) - log5(1-x) = log5(2x+3)
ОДЗ {x+1>0⇒x>-1 {1-x>0⇒x<1<br>{2x+3>0⇒x>-1.5 x∈(-1;1) log(5)[(x+1)/(1-x)]=log(5)(2x+3) (x+1)/(1-x)=2x+3 (2x+3)(1-x)-(x+1)=0 2x-2x²+3-3x-x-1=0 2x²+2x-2=0 x²+x-1=0 D=1+4=5 x1=(-1-√5)/2∉ОДЗ х2=(-1+√5)/2