Представьте в виде дроби (4-3b)/(b^2-2b)+3/(b-2)

Ответ:

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

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

\[\frac{x - 6y^{2}}{2y} + 3y^{\backslash 2y} = \frac{x - 6y^{2} + 6y^{2}}{2y} = \frac{x}{2y}\]

\[при\ x = - 8\ \ и\ \ y = 0,1:\]

\[\frac{x}{2y} = \frac{- 8}{2 \cdot 0,1} = \frac{- 8}{0,2} = - 40.\]

\[\frac{2}{x - 4} - \frac{x + 8}{x^{2} - 16} - \frac{1}{x} = \frac{2^{\backslash x(x + 4)}}{x - 4} -\]

\[- \frac{x + 8^{\backslash x}}{(x - 4)(x + 4)} - \frac{1^{\backslash x^{2} - 16}}{x} =\]

\[= \frac{2x(x + 4) - x(x + 8) - \left( x^{2} - 16 \right)}{x(x - 4)(x + 4)} =\]

\[= \frac{2x^{2} + 8x - x^{2} - 8x - x^{2} + 16}{x(x - 4)(x + 4)} =\]

\[= \frac{16}{x(x^{2} - 16)}\]