log5(x+1) - log5(1-x) = log5(2x+3)

ОДЗ{x+1>0⇒x>-1{1-x>0⇒x<1{2x+3>0⇒x>-1.5x∈(-1;1)log(5)[(x+1)/(1-x)]=log(5)(2x+3)(x+1)/(1-x)=2x+3(2x+3)(1-x)-(x+1)=02x-2x²+3-3x-x-1=02x²+2x-2=0x²+x-1=0D=1+4=5x1=(-1-√5)/2∉ОДЗх2=(-1+√5)/2