Докажите тождество ((2a-0,5b)/(4a^2+ab+0,25b^2)+24ab/(64a^3-b^3)+1/(2a-0,5b))*(4a-b)/4=1.

\[Упростим\ левую\ часть\ \]

\[равенства:\]

\[1)\ \frac{2a - 0,5b}{4a^{3} + ab + 0,25b^{2}} =\]

\[= \frac{0,5 \cdot (4a - b)}{0,25 \cdot \left( 16a^{2} + 4ab + b^{2} \right)} =\]

\[= \frac{2 \cdot (4a - b)}{16a^{2} + 4ab + b^{2}}\]

\[2.\ \ \frac{1}{2a - 0,5b} = \frac{1}{0,5 \cdot (4a - b)} =\]

\[= \frac{2}{4a - b}\]

\[= \frac{32a^{2} - 16ab + 4b^{2}}{(4a - b)\left( 16a^{2} + 4ab + b^{2} \right)} =\]

\[= \frac{64a^{2} + 16ab + 4b^{2}}{(4a - b)(16a^{2} + 4ab + b^{2})} =\]

\[= \frac{4 \cdot \left( 16a^{2} + 4ab + b^{2} \right)}{(4a - b)\left( 16a^{2} + 4ab + b^{2} \right)} =\]

\[= \frac{4}{4a - b}\]

\[4.\frac{4}{4a - b} \cdot \frac{4a - b}{4} = 1\]

\[1 = 1 \Longrightarrow ч.т.д.\]