Логарифмы. нужно упростить. даю много

$(\frac{log_ab+log_ac}{log_ab-log_ac} +1)*(\frac{log_ab-log_ac}{log_ab+log_ac} +1):\frac{4log_ab}{log^2_ab-log^2_ac} =$ $\frac{log_ab+log_ac+log_ab-log_ac}{log_ab-log_ac} *\frac{log_ab-log_ac+log_ab+log_ac}{log_ab+log_ac} *\frac{log^2_ab-log^2_ac}{4log_ab} =$ $\frac{log_ab+log_ab}{log_ab-log_ac} *\frac{log_ab+log_ab}{log_ab+log_ac} *\frac{log^2_ab-log^2_ac}{4log_ab} =\frac{2log_ab}{log_ab-log_ac} *\frac{2log_ab}{log_ab+log_ac} *\frac{log^2_ab-log^2_ac}{4log_ab} =$ $=\frac{4log^2_ab}{(log_ab-log_ac)(log_ab+log_ac)}*\frac{log^2_ab-log^2_ac}{4log_ab} =\frac{4log^2_ab}{log^2_ab-log^2_ac}*\frac{log^2_ab-log^2_ac}{4log_ab} =log_ab$

$\frac{a^{log_ab+1}*b^{log_ba-2}}{a^{log_ab+2}*b^{log_ba-1}} =\frac{a^{log_ab}*a*b^{log_ba}*b^{-2}}{a^{log_ab}*a^2*b^{log_ba}*b^{-1}}=\frac{b*a*a*b^{-2}}{b*a^2*a*b^{-1}}=\frac{a^2*b^{-1}}{a^3}= \frac{b^{-1}}{a} =\frac{1}{ab}$