\[\frac{5}{x + 3} - \frac{4}{x - 1}\]
\[x + 3
eq 0\ \ \ \ \ \ \ x - 1
eq 0\]
\[x
eq - 3\ \ \ \ \ \ \ \ \ \ \ x
eq 1\]
\[Ответ:x
eq - 3;\ \ x
eq 1.\]
\[\frac{18a^{4}b^{8}}{6a^{7}b^{4}} = \frac{3b^{4}}{a^{3}}\]
\[\frac{2x^{2} + xy}{2xy + y^{2}} = \frac{x(2x + y)}{y(2x + y)} = \frac{x}{y}\]
\[\frac{a}{a - 3} - \frac{a^{2} - 2a + 6}{a^{2} - 3a} =\]
\[= \frac{a^{\backslash a}}{a - 3} - \frac{a^{2} - 2a + 6}{a(a - 3)} =\]
\[= \frac{a^{2} - a^{2} + 2a - 6}{a(a - 3)} = \frac{2a - 6}{a(a - 3)} =\]
\[= \frac{2(a - 3)}{a(a - 3)} = \frac{2}{a}\]
\[\frac{a^{2} + 3b}{a} - a^{\backslash a} = \frac{a^{2} + 3b - a^{2}}{a} = \frac{3b}{a}\text{\ \ }\]
\[при\ a = 0,6;\ \ b = 2:\]
\[\frac{3b}{a} = \frac{3 \cdot 2}{0,6} = \frac{6}{0,6} = \frac{60}{6} = 10.\]
\[y = \frac{x^{2} - 6x + 9}{3 - x} = \frac{(3 - x)^{2}}{3 - x} = 3 - x;\ \ \ x
eq 3\]