1+log2 (x+1)=log2 (7x+2) -log2 (x-1)

ОДЗ: x+1>0 => x>-17x+2> => x>-2/7x-1>0 => x>1log2 (2)+log2 (x+1)=log2 (7x+2) -log2 (x-1) log2 (2x+2)=log2 (7x+2/x-1) 2x+2=7x+2/x-1 (2x+2) (x-1)=7x+22x^2-2x+2x-2=7x+22x^2-7x-4=0D=49+32=81x (1)=7-9/4=-0,5 (не удовлетворяет ОДЗ) x (2)=7+9/4=4