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

\[\frac{3a^{2} - 6ab + 3b^{2}}{4(a - b)(a + b)} =\]

\[= \frac{3\left( a^{2} - 2ab + b^{2} \right)}{4(a - b)(a + b)} =\]

\[= \frac{3(a - b)^{2}}{4(a - b)(a + b)} = \frac{3(a - b)}{4(a + b)};\]

\[a = 2;\ \ b = 1:\]

\[\frac{3(2 - 1)}{4(2 + 1)} = \frac{3}{12} = \frac{1}{4} = 0,25.\]