Question

Решите ,пожалуйста, если можно с обьяснением

Answer 1

1)

$\displaystyle \frac{(log_ab+log_ba+1)(log_aa-log_ab)}{1-log^3_ab}=\\\\=\frac{(log_ab+\frac{1}{log_ab}+1)(1-log_ab)}{1-log^3_ab}=\\\\=\frac{(log^2_ab+log_ab+1)*\frac{1}{log_ab}*(1-log_ab) }{1-log^3_ab}=\\\\=\frac{(1-log^3_ab)*\frac{1}{log_ab} }{(1-log^3_ab)}=\frac{1}{log_ab}$

2)

$\displaystyle \frac{1+log^3_xy}{(\frac{1}{log_xy}+log_xy-1)(log_xx+log_xy)}=\\\\=\frac{(1+log^3_xy)}{(1+log^2_xy-log_xy)*(1+log_xy)*\frac{1}{log_xy} }=log_xy$