Найдите значение выражения ((a-3b)^2/(2*(6a-b))+(a+3b)^2/(3*(b-6a))+(14ab-b^2)/(3*(6a-b)) при a=1/5; b=-4/5.

Ответ:

\[= \frac{a^{2} + 2ab + b^{2}}{6(6a - b)} = \frac{(a + b)^{2}}{36a - 6b}\]

\[a = \frac{1}{5};\ \ b = - \frac{4}{5}:\]

\[\frac{\left( \frac{1}{5} - \frac{4}{5} \right)^{2}}{36 \cdot \frac{1}{5} - 6 \cdot \frac{4}{5}} = \frac{\left( - \frac{3}{5} \right)^{2}}{\frac{36}{5} - \frac{24}{5}} =\]

\[= \frac{9}{25}\ :\frac{12}{5} = \frac{9}{25} \cdot \frac{5}{12} = \frac{3}{5 \cdot 4} =\]

\[= \frac{3}{20} = 0,15.\]