Постройте график функции y=(x^2-4x+4)/(2-x).

\[\frac{5}{x + 3} - \frac{4}{x - 1}\]

\[x + 3 \neq 0\ \ \ \ \ \ \ x - 1 \neq 0\]

\[x \neq - 3\ \ \ \ \ \ \ \ \ \ \ x \neq 1\]

\[Ответ:x \neq - 3;\ \ x \neq 1.\]

\[\frac{18a^{4}b^{8}}{6a^{7}b^{4}} = \frac{3b^{4}}{a^{3}}\]

\[\frac{2x^{2} + xy}{2xy + y^{2}} = \frac{x(2x + y)}{y(2x + y)} = \frac{x}{y}\]

\[\frac{a}{a - 3} - \frac{a^{2} - 2a + 6}{a^{2} - 3a} =\]

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

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

\[= \frac{2(a - 3)}{a(a - 3)} = \frac{2}{a}\]

\[\frac{a^{2} + 3b}{a} - a^{\backslash a} = \frac{a^{2} + 3b - a^{2}}{a} = \frac{3b}{a}\text{\ \ }\]

\[при\ a = 0,6;\ \ b = 2:\]

\[\frac{3b}{a} = \frac{3 \cdot 2}{0,6} = \frac{6}{0,6} = \frac{60}{6} = 10.\]

\[y = \frac{x^{2} - 6x + 9}{3 - x} = \frac{(3 - x)^{2}}{3 - x} = 3 - x;\ \ \ x \neq 3\]