Question

логарифмы. нужно упростить. даю много

Answer 1

чтобы проще работать введем обозначения log b поосн.а=х, log c по осн а=у, тогда(x+y/x-y   +1)*(x-y/x+y  +1)*(x-y)(x+y)/4x=x+y+x-y/x-y +x+y/x+y *(x-y)(x+y)/4x=4x^/4x=x=logb по осн.а

(b+1)(a-2)/(b+2)(a-1) это ответ ко 2-му

Answer 2

1)

$(\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$

2)

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