Найдите допустимые значения переменной в выражении

\[\frac{3x - 6}{x - 2} + \frac{2x - 6}{x + 1}\]

\[x - 2 \neq 0\ \ \ \ \ \ \ \ x + 1 \neq 0\]

\[x \neq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq - 1\]

\[Ответ:x \neq - 1;\ \ x \neq 2.\]

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

\[\frac{a + 3}{a^{2} + a} - \frac{1}{a + 1} + \frac{2}{a} =\]

\[= \frac{a + 3}{a(a + 1)} - \frac{1^{\backslash a}}{a + 1} + \frac{2^{\backslash a + 1}}{a} =\]

\[= \frac{a + 3 - a + 2a + 2}{a(a + 1)} = \frac{2a + 5}{a(a + 1)}\]

\[\frac{2x^{2} - 4x + 7}{x - 2} = \frac{2 \cdot \left( x^{2} - 4 \right) + 7}{x - 2} =\]

\[= \frac{2x \cdot (x - 2) + 7}{x - 2} = \frac{2x(x - 2)}{x - 2} + \frac{7}{x - 2} =\]

\[= 2x + \frac{7}{x - 2}\]

\[y = \frac{x^{2} - 6x + 9}{3 - x} + \frac{4x^{2} - 6x}{x} =\]

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

\[= 3 - x + 4x - 6 = 3x - 3\]

\[y = 3x - 3;\ \ \ x \neq 0;\ \ x \neq 3\]

\[\frac{ax^{2} + x + b}{x + 2} = 2x - 3;\ \ \ \ \ \ x \neq - 2\]

\[ax^{2} + x + b = (2x - 3)(x + 2)\]

\[ax^{2} + x + b = 2x^{2} - 3x + 4x - 6\]

\[ax^{2} + x + b = 2x^{2} + x - 6\]

\[a = 2;\ \ \ b = - 6.\]

\[Ответ:a = 2;\ \ b = - 6.\]

\[\frac{5x + 15}{x + 3} + \frac{3x - 1}{x - 2}\]

\[x + 3 \neq 0\ \ \ \ \ \ \ x - 2 \neq 0\]

\[x \neq - 3\ \ \ \ \ \ \ \ \ \ \ x \neq 2\]

\[Ответ:x \neq - 3;x \neq 2.\]

\[\frac{9 - x²}{x^{2} + 6x + 9} = \frac{(3 - x)(3 + x)}{(x + 3)^{2}} = \frac{3 - x}{3 + x}\]

\[\frac{a - 3}{a^{2} - a} - \frac{1}{a - 1} - \frac{4}{a} =\]

\[= \frac{a - 3}{a(a - 1)} - \frac{1^{\backslash a}}{a - 1} - \frac{4^{\backslash a - 1}}{a} =\]

\[= \frac{a - 3 - a - 4a + 4}{a(a - 1)} = \frac{1 - 4a}{a(a - 1)}\]

\[\frac{3x^{2} - 12x + 5}{x - 4} = \frac{3x(x - 4) + 5}{x - 4} =\]

\[= \frac{3x(x - 4)}{x - 4} + \frac{5}{x - 4} = 3x + \frac{5}{x - 4}\]

\[y = \frac{x^{2} - 4x + 4}{2 - x} + \frac{3x^{2} - 4x}{x} =\]

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

\[= 2x - 2\]

\[y = 2x - 2;\ \ \ x \neq 0;\ \ x \neq 2.\]

\[\frac{ax^{2} - 8x + b}{x - 3} = 3x + 1;\ \ \ \ \ x \neq 3\]

\[ax^{2} - 8x + b = (3x + 1)(x - 3)\]

\[ax^{2} - 8x + b = 3x^{2} + x - 9x - 3\]

\[ax^{2} - 8x + b = 3x^{2} - 8x - 3\]

\[a = 3;\ \ \ b = - 3.\]

\[Ответ:a = 3;\ \ b = - 3.\]

\[\frac{4x + 4}{x + 1} + \frac{x^{2} - 6x + 9}{3 - x} + \frac{2x + 5}{4 - x^{2}}\]

\[x + 1 \neq 0;\ \ \ \ \ 3 - x \neq 0;\ \ \ \ \ 4 - x^{2} \neq 0\]

\[x \neq - 1;\ \ \ \ \ \ \ \ \ x \neq 3;\ \ \ \ \ \ \ \ \ \ \ \ \ x \neq \pm 2\]

\[Ответ:x \neq - 1;\ \ \ x \neq 3;\ \ x \neq \pm 2.\]

\[\frac{ax + 2b + 2a + bx}{3a - bx + 3b - ax} =\]

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

\[= \frac{x + 2}{3 - x}\]

\[\frac{a^{2}}{a^{2} - 9} + \frac{3}{a + 3} - \frac{a}{a - 3} =\]

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

\[= \frac{a^{2} + 3a - 9 - a^{2} - 3a}{(a - 3)(a + 3)} =\]

\[= - \frac{9}{a^{2} - 9} = \frac{9}{9 - a²}\]

\[при\ a = 1,5:\]

\[\frac{9}{9 - a^{2}} = \frac{9}{(3 - 1,5)(3 + 1,5)} = \frac{9}{1,5 \cdot 4,5} =\]

\[= \frac{2}{1,5} = \frac{20}{15} = \frac{4}{3}.\]

\[Ответ:\ \frac{4}{3}.\]

\[y = \frac{x^{2} - 4x + 4}{|x - 2|} = \frac{(x - 2)^{2}\ }{|x - 2|} =\]

\[= \frac{|x - 2|^{2}}{|x - 2|} = |x - 2|\]

\[y = |x - 2|;\ \ \ x \neq 2.\]

\[\frac{x^{3} + x^{2} + x - 3}{x - 1}\]

\[Разложим\ числитель\ на\ множители:\]

\[x^{3} + x^{2} + x - 3 =\]

\[= \left( x^{3} - x^{2} \right) + \left( 2x^{2} - 2x \right) + (3x - 3) =\]

\[= x^{2}(x - 1) + 2x(x - 1) + 3(x - 1) =\]

\[= (x - 1)\left( x^{2} + 2x + 3 \right)\]

\[Подставим:\]

\[\frac{(x - 1)(x^{2} + 2x + 3)}{x - 1} = x^{2} + 2x + 3.\]

\[\frac{x + 3}{6x^{2} + x - 2} = \frac{a^{\backslash 3x + 2}}{2x - 1} + \frac{b^{\backslash 2x - 1}}{3x + 2}\]

\[\frac{x + 3}{6x^{2} + x - 2} = \frac{3ax + 2a + 2bx - b}{(2x - 1)(3x + 2)}\]

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

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

\[\left\{ \begin{matrix} 3a + 2b = 1\ \ \ \ \ \ \ \\ 2a - b = 3\ \ | \cdot 2\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} 3a + 2b = 1 \\ 4a - 2b = 6\ \\ \end{matrix}( + ) \right.\ \]

\[7a = 7\]

\[a = 1;\]

\[b = 2a - 3 = 2 - 3 = - 1.\]

\[\left\{ \begin{matrix} a = 1\ \ \ \\ b = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:a = 1;\ \ b = - 1.\]

\[\frac{3x - 3}{x - 1} + \frac{x^{2} + 4x + 4}{x + 2} - \frac{5x - 4}{9 - x^{2}}\]

\[x - 1 \neq 0;\ \ x + 2 \neq 0;\ \ \ 9 - x^{2} \neq 0\]

\[x \neq 1;\ \ \ \ \ \ \ \ \ \ x \neq - 2;\ \ \ \ \ \ \ \ x \neq \pm 3\]

\[Ответ:x \neq - 2;\ \ x \neq 1;\ \ x \neq \pm 3.\]

\[\frac{ax + 3b - 3a - bx}{ax - b + a - bx} =\]

\[= \frac{(ax - bx) - (3a - 3b)}{(ax - bx) + (a - b)} =\]

\[= \frac{x(a - b) - 3(a - b)}{x(a - b) + (a - b)} =\]

\[= \frac{(a - b)(x - 3)}{(a - b)(x + 1)} = \frac{x - 3}{x + 1}\]

\[\frac{4^{\backslash x - 2}}{x + 2} - \frac{3^{\backslash x + 2}}{x - 2} + \frac{12}{x^{2} - 4} =\]

\[= \frac{4x - 8 - 3x - 6 + 12}{(x - 2)(x + 2)} =\]

\[= \frac{x - 2}{(x - 2)(x + 2)} = \frac{1}{x + 2}\]

\[при\ x = - 0,6:\]

\[\frac{1}{x + 2} = \frac{1}{- 0,6 + 2} = \frac{1}{1,4} = \frac{10}{14} = \frac{5}{7}.\]

\[Ответ:\ \frac{5}{7}.\]

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

\[= |x - 3|;\ \ \ \ x \neq 3\]

\[\frac{x^{3} - 2x^{2} + x + 4}{x + 1}\]

\[Разложим\ числитель\ на\ множители:\]

\[x^{2} - 2x^{2} + x + 4 =\]

\[= \left( x^{3} + x^{2} \right) + \left( - 3x^{2} - 3x \right) + (4x + 4) =\]

\[= x^{2}(x + 1) - 3x(x + 1) + 4(x + 1) =\]

\[= (x + 1)\left( x^{2} - 3x + 4 \right)\]

\[Подставим:\]

\[\frac{(x + 1)\left( x^{2} - 3x + 4 \right)}{x + 1} = x^{2} - 3x + 4.\]

\[\frac{8x + 1}{6x^{2} + 7x - 3} = \frac{a^{\backslash 3x - 1}}{2x + 3} + \frac{b^{\backslash 2x + 3}}{3x - 1}\]

\[\frac{8x + 1}{6x^{2} + 7x - 3} = \frac{3ax - a + 2bx + 3b}{(2x + 3)(3x - 1)}\]

\[\frac{8x + 1}{6x^{2} + 7x - 3} = \frac{(3a + 2b)x + (3b - a)}{6x^{2} + 9x - 2x - 3}\]

\[\frac{8x + 1}{6x^{2} + 7x - 3} = \frac{(3a + 2b)x + (3b - a)}{6x^{2} + 7x - 3}\]

\[\left\{ \begin{matrix} 3a + 2b = 8 \\ 3b - a = 1\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} a = 3b - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3(3b - 1) + 2b = 8\ \\ \end{matrix} \right.\ \]

\[9b - 3 + 2b = 8\]

\[11b = 11\]

\[b = 1.\]

\[a = 3b - 1 = 3 - 1 = 2.\]

\[\left\{ \begin{matrix} a = 2 \\ b = 1 \\ \end{matrix} \right.\ \]

\[Ответ:a = 2;\ \ b = 1.\]

\[\frac{a - 3}{a^{2} + 6a} = 0\]

\[a - 3 = 0\]

\[a = 3.\]

\[Допустимые\ значения:\]

\[a^{2} + 6a \neq 0\]

\[a(a + 6) \neq 0\]

\[a \neq 0;\ \ \ \ a + 6 \neq 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a \neq - 6.\]

\[\frac{6y - 3x}{x^{2} - 4y^{2}} = \frac{3(2y - x)}{(x - 2y)(x + 2y)} = - \frac{3}{x + 2y}\ \]

\[при\ x = 0,2;\ \ y = 0,4:\]

\[- \frac{3}{x + 2y} = - \frac{3}{0,2 + 2 \cdot 0,4} = - \frac{3}{0,2 + 0,8} =\]

\[= - \frac{3}{1} = - 3.\]

\[\left( 2^{\backslash a + 1} + \frac{a}{a + 1} \right)\ :\frac{12a + 8}{3a^{2} + 3a} =\]

\[= \frac{2a + 2 + a}{a + 1} \cdot \frac{3a^{2} + 3a}{12a + 8} =\]

\[= \frac{3a + 2}{a + 1} \cdot \frac{3a(a + 1)}{4(3a + 2)} = \frac{3a}{4}\]

\[\frac{a}{b} = 3 \Longrightarrow a = 3b:\]

\[\frac{2a + 3b}{3a + 2b} = \frac{2 \cdot 3b + 3b}{3 \cdot 3b + 2b} = \frac{9b}{11b} = \frac{9}{11}.\]

\[A = \frac{2n^{2} + 3n + 5}{n} = 2n + 3 + \frac{5}{n}\]

\[n = 1 \Longrightarrow A = 2 \cdot 1 + 3 + \frac{5}{1} = 10.\]

\[n = - 1 \Longrightarrow A = - 2 + 3 - 5 = - 4.\]

\[n = 5 \Longrightarrow A = 10 + 3 + 1 = 14.\]

\[n = - 5 \Longrightarrow A = - 10 + 3 - 1 = - 8.\]

\[y = \frac{x - 3}{x^{2} - 3x} = \frac{x - 3}{x(x - 3)} = \frac{1}{x}\]

\[y = \frac{1}{x};\ \ x \neq 0;\ \ x \neq 3\]

\[y < 0\ при\ x < 0.\]

\[\frac{4 + a}{a^{2} - 3a} = 0\]

\[4 + a = 0\]

\[a = - 4.\]

\[Допустимые\ значения:\]

\[a^{2} - 3a \neq 0\]

\[a(a - 3) \neq 0\]

\[a \neq 0;\ \ a \neq 3.\]

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

\[при\ x = 0,3;y = - 0,35:\]

\[\frac{4}{x - 2y} = \frac{4}{0,3 - 2 \cdot ( - 0,35)} = \frac{4}{0,3 + 0,7} =\]

\[= \frac{4}{1} = 4.\]

\[\left( \frac{2a}{2a - 1} + 1^{\backslash 2a - 1} \right)\ :\frac{4a^{2} - a}{6a - 3} =\]

\[= \frac{2a + 2a - 1}{2a - 1} \cdot \frac{6a - 3}{4a^{2} - a} =\]

\[= \frac{(4a - 1) \cdot 3(2a - 1)}{(2a - 1) \cdot a(4a - 1)} = \frac{3}{a}.\]

\[\frac{a}{b} = 2 \Longrightarrow a = 2b:\]

\[\frac{4a + 3b}{3a + 4b} = \frac{4 \cdot 2b + 3b}{3 \cdot 2b + 4b} = \frac{11b}{10b} = \frac{11}{10}.\]

\[A = \frac{3n^{2} - 2n + 3}{n} = 3n - 2 + \frac{3}{n}\]

\[n = 1 \Longrightarrow A = 3 - 2 + 3 = 4.\]

\[n = - 1 \Longrightarrow A = - 3 - 2 - 3 = - 8.\]

\[n = 3 \Longrightarrow A = 9 - 2 + 1 = 8.\]

\[n = - 3 \Longrightarrow A = - 9 - 2 - 1 = - 12.\]

\[y = \frac{x + 2}{x^{2} + 2x} = \frac{x + 2}{x(x + 2)} = \frac{1}{x}\]

\[y = \frac{1}{x};\ \ \ x \neq 0;\ \ x \neq - 2\]

\[y > 0\ при\ x > 0.\]

\[\frac{a^{2} - 2a}{a^{2} - a - 2} = 0\]

\[a^{2} - 2a = 0\]

\[a(a - 2) = 0\]

\[a = 0;\ \ \ \]

\[a = 2\ (не\ подходит).\]

\[Допустимые\ значения:\]

\[a^{2} - a - 2 \neq 0\]

\[a^{2} - 2a + a - 2 \neq 0\]

\[a(a - 2) + (a - 2) \neq 0\]

\[(a - 2)(a + 1) \neq 0\]

\[a \neq 2;\ \ a \neq - 1.\]

\[\frac{ax - ay - bx + by}{ax - bx + 2ay - 2by} =\]

\[= \frac{a(x - y) - b(x - y)}{x(a - b) + 2y(a - b)} =\]

\[= \frac{(x - y)(a - b)}{(a - b)(x + 2y)} = \frac{x - y}{x + 2y}\]

\[при\ x = 1,2;y = - 0,1:\]

\[\frac{x - y}{x + 2y} = \frac{1,2 + 0,1}{1,2 - 0,2} = \frac{1,3}{1} = 1,3.\]

\[\left( \frac{a + b^{\backslash a + b}}{a - b} - \frac{a - b^{\backslash a - b}}{a + b} \right)\ :\frac{\text{ab}}{a^{2} - b^{2}} =\]

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

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

\[\frac{2a - b}{a + b} = 1:\]

\[2a - b = a + b\]

\[2a - a = b + b\]

\[a = 2b.\]

\[\frac{3a - 4b}{a + 2b} = \frac{3 \cdot 2b - 4b}{2b + 2b} = \frac{2b}{4b} = \frac{1}{2}.\]

\[A = \frac{n^{2} + n + 3}{n + 2} = \frac{n^{2} + 2n - n - 2 + 5}{n + 2} =\]

\[= \frac{n(n + 2) - (n + 2) + 5}{n + 2} =\]

\[= \frac{(n + 2)(n - 1) + 5}{n + 2} = n - 1 + \frac{5}{n + 2}\]

\[n + 2\ должно\ быть\ кратно\ 5,\ то\ есть\ \]

\[равно\ 1;\ - 1;5;\ - 5.\]

\[n + 2 = 1\]

\[n = - 1 \Longrightarrow A = - 1 - 1 + 5 = 3.\]

\[n + 2 = - 1\]

\[n = - 3 \Longrightarrow A = - 3 - 1 - 5 = - 9.\]

\[n + 2 = 5\]

\[n = 3 \Longrightarrow A = 3 - 1 + 1 = 3.\]

\[n + 2 = - 5\]

\[n = - 7 \Longrightarrow A = - 7 - 1 - 1 = - 9.\]

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

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

\(y = \frac{4}{x};\ \ \ x \neq 0.\)

\[y < 0\ при\ x < 0.\]

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

\[Допустимые\ значения:\]

\[a^{2} + 2a - 3 \neq 0\]

\[a^{2} + 3a - a - 3 \neq 0\]

\[a(a + 3) - (a + 3) \neq 0\]

\[(a + 3)(a - 1) \neq 0\]

\[a \neq - 3;\ \ a \neq 1.\]

\[\frac{a^{2} + 3a}{a^{2} + 2a - 3} = 0\]

\[a^{2} + 3a = 0\]

\[a(a + 3) = 0\]

\[a = 0;\]

\[a = - 3\ (не\ подходит).\]

\[\frac{ax - 2ay + bx - 2by}{ax + bx + ay + by} =\]

\[= \frac{x(a + b) - 2y(a + b)}{x(a + b) + y(a + b)} =\]

\[= \frac{(a + b)(x - 2y)}{(a + b)(x + y)} = \frac{x - 2y}{x + y}\]

\[при\ x = 1,3;\ \ y = - 0,3:\]

\[\frac{x - 2y}{x + y} = \frac{1,3 - 2 \cdot ( - 0,3)}{1,3 - 0,3} =\]

\[= \frac{1,3 + 0,6}{1} = 1,9.\]

\[\left( \frac{a - b^{\backslash a - b}}{a + b} - \frac{a + b^{\backslash a + b}}{a - b} \right)\ :\frac{b^{2}}{a^{2} - b^{2}} =\]

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

\[= \frac{- 4ab \cdot \left( a^{2} - b^{2} \right)}{\left( a^{2} - b^{2} \right) \cdot b²} = - \frac{4a}{b}.\]

\[\frac{3a - 5b}{a - b} = 1:\]

\[3a - 5b = a - b\]

\[3a - a = 5b - b\]

\[2a = 4b\]

\[a = 2b.\]

\[\frac{2a - 3b}{2a + b} = \frac{2 \cdot 2b - 3b}{2 \cdot 2b + b} = \frac{b}{5b} = \frac{1}{5}.\]

\[A = \frac{n^{2} + 2n + 2}{n + 3} = \frac{n^{2} + 3n - n - 3 + 5}{n + 3} =\]

\[= \frac{n(n + 3) - (n + 3) + 5}{n + 3} =\]

\[= \frac{(n + 3)(n - 1) + 5}{n + 3} = n - 1 + \frac{5}{n + 3}\]

\[n + 3\ должно\ быть\ кратно\ 5,\ то\ есть\ \]

\[равно\ 1;\ - 1;5;\ - 5.\]

\[n + 3 = 1\]

\[n = - 2 \Longrightarrow A = - 2 - 1 + 5 = 2.\]

\[n + 3 = - 1\]

\[n = - 4 \Longrightarrow A = - 4 - 1 - 5 = - 10.\]

\[n + 3 = 5\]

\[n = 2 \Longrightarrow A = 2 - 1 + 1 = 2.\]

\[n + 3 = - 5\]

\[n = - 8 \Longrightarrow A = - 8 - 1 - 1 = - 10.\]

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

\[= \frac{x^{2} - 6x + 9 - x^{2} - 6x - 9}{6x^{2}} =\]

\[= \frac{- 12x}{6x²} = - \frac{2}{x}\]

\[y = - \frac{2}{x};\ \ \ x \neq 0\]

\[y < при\ x > 0.\]

\[\frac{a^{3} - 4a}{a^{2} - a - 2}\]

\[Допустимые\ значения:\]

\[a^{2} - a - 2 \neq 0\]

\[a^{2} - 2a + a - 2 \neq 0\]

\[a(a - 2) + (a - 2) \neq 0\]

\[(a - 2)(a + 1) \neq 0\]

\[a \neq 2;\ \ \ a \neq - 1.\]

\[\frac{a^{3} - 4a}{a^{2} - a - 2} = 0\]

\[a^{3} - 4a = 0\]

\[a\left( a^{2} - 4 \right) = 0\]

\[a(a - 2)(a + 2) = 0\]

\[a = 0;\ \ \]

\[a = 2\ (не\ подходит);\]

\[a = - 2.\]

\[\frac{a^{2} - ac + b^{2} + 2ab - bc}{ab + ac + b^{2} - c^{2}} =\]

\[= \frac{\left( a^{2} + 2ab + b^{2} \right) - (ac + bc)}{(ab + ac) + \left( b^{2} - c^{2} \right)} =\]

\[= \frac{(a + b)(a + b) - c(a + b)}{a(b + c) + (b - c)(b + c)} =\]

\[= \frac{(a + b)(a + b - c)}{(b + c)(a + b - c)} = \frac{a + b}{b + c}\]

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

\[= \left( \frac{a^{\backslash a + b}}{a - b} - \frac{a^{\backslash a - b}}{a + b} + \frac{2b^{2}}{(a - b)(a + b)} \right) \cdot \frac{a - b}{(a + b)^{2}} =\]

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

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

\[= \frac{2b(a + b)}{(a + b)(a + b)^{2}} = \frac{2b}{(a + b)^{2}}\]

\[\frac{3a + b}{a + 2b} = 2:\]

\[3a + b = 2(a + 2b)\]

\[3a + b = 2a + 4b\]

\[3a - 2a = 4b - b\]

\[a = 3b.\]

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

\[= \frac{2 \cdot 9b^{2} - 3b \cdot b + b^{2}}{3 \cdot 9b^{2} - 2 \cdot 3b \cdot b + b^{2}} =\]

\[= \frac{18b^{2} - 3b^{2} + b^{2}}{27b^{2} - 6b^{2} + b^{2}} = \frac{16b^{2}}{22b^{2}} = \frac{8}{11}\]

\[xy + 3x - 2y = 9\]

\[x(y + 3) - 2y = 9\]

\[x(y + 3) = 9 + 2y\]

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

\[= \frac{2(y + 3) + 3}{y + 3} = 2 + \frac{3}{y + 3}\]

\[y + 3\ должно\ быть\ делителем\ числа\ 3,\]

\[то\ есть\ быть\ равно\ 1;\ - 1;3;\ - 3.\]

\[1)\ y + 3 = 1\]

\[y = - 2 \Longrightarrow x = 2 + 3 = 5.\]

\[2)\ y + 3 = - 1\]

\[y = - 4 \Longrightarrow x = 2 - 3 = - 1.\]

\[3)\ y + 3 = 3\]

\[y = 0 \Longrightarrow x = 2 + 1 = 3.\]

\[4)\ y + 3 = - 3\]

\[y = - 6 \Longrightarrow x = 2 - 1 = 1.\]

\[y = \frac{18 - 12x}{x^{2} - 3x} - \frac{6}{3 - x} =\]

\[= \frac{18 - 12x}{x(x - 3)} + \frac{6^{\backslash x}}{x - 3} = \frac{18 - 12x + 6x}{x(x - 3)} =\]

\[= \frac{18 - 6x}{x(x - 3)} = \frac{- 6(x - 3)}{x(x - 3)} = - \frac{6}{x}\]

\[y = - \frac{6}{x};\ \ \ x \neq 0;\ \ \ x \neq 3\]

\[\frac{a^{3} - 9a}{a^{2} + 2a - 3}\]

\[Допустимые\ значения:\]

\[a^{2} + 2a - 3 \neq 0\]

\[a^{2} + 3a - a - 3 \neq 0\]

\[a(a + 3) - (a + 3) \neq 0\]

\[(a + 3)(a - 1) \neq 0\]

\[a \neq - 3;\ \ \ a \neq 1.\]

\[\frac{a^{3} - 9a}{a^{2} + 2a - 3} = 0\]

\[a^{3} - 9a = 0\]

\[a\left( a^{2} - 9 \right) = 0\]

\[a(a - 3)(a + 3) = 0\]

\[a = 0;\ \]

\[a = 3;\]

\[a = - 3\ (н\ подходит).\]

\[\frac{a^{2} + ac + b^{2} - 2ab - bc}{ab + 2bc - ac - b^{2} - c^{2}} =\]

\[= \frac{\left( a^{2} - 2ab + b^{2} \right) + (ac - bc)}{(ab - ac) - \left( b^{2} - 2bc + c^{2} \right)} =\]

\[= \frac{(a - b)^{2} + c(a - b)}{a(b - c) - (b - c)^{2}} =\]

\[= \frac{(a - b)(a - b + c)}{(b - c)(a - b + c)} = \frac{a - b}{b - c}\]

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

\[= \left( \frac{a^{\backslash a - b}}{a + b} - \frac{b^{2} + a^{2}}{(a - b)(a + b)} - \frac{a^{\backslash a + b}}{a - b} \right) \cdot \frac{2}{a + b} =\]

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

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

\[= \frac{- (a + b)^{2} \cdot 2}{(a - b)(a + b)(a + b)} =\]

\[= - \frac{2}{a - b} = \frac{2}{b - a}\]

\[\frac{a + 4b}{2a - b} = 2:\]

\[a + 4b = 2(2a - b)\]

\[a + 4b = 4a - 2b\]

\[a - 4a = - 2b - 4b\]

\[- 3a = - 6b\]

\[a = 2b.\]

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

\[= \frac{4b^{2} - 4b^{2} + 3b^{2}}{8b^{2} + 2b^{2} + b^{2}} = \frac{3b^{2}}{11b^{2}} = \frac{3}{11}\]

\[xy + 3y - x = 6\]

\[x(y - 1) + 3y = 6\]

\[x(y - 1) = 6 - 3y\]

\[x = \frac{6 - 3y}{y - 1} = \frac{( - 3y + 3) + 3}{y - 1} =\]

\[= \frac{- 3(y - 1) + 3}{y - 1} = - 3 + \frac{3}{y - 1}\]

\[y - 1\ должно\ быть\ делителем\ числа\ 3,\]

\[то\ есть\ быть\ равно\ 1;\ - 1;3;\ - 3.\]

\[1)\ y - 1 = 1\]

\[y = 2 \Longrightarrow x = - 3 + 3 = 0.\]

\[2)\ y - 1 = - 1\]

\[y = 0 \Longrightarrow x = - 3 - 3 = - 6\]

\[3)\ y - 1 = 3\]

\[y = 4 \Longrightarrow x = - 3 + 1 = 2.\]

\[4)\ y - 1 = - 3\]

\[y = - 2 \Longrightarrow x = - 3 - 1 = - 4.\]

\[y = \frac{6}{x + 2} - \frac{2x - 8}{x^{2} + 2x} = \frac{6^{\backslash x}}{x + 2} - \frac{2x - 8}{x(x + 2)} =\]

\[= \frac{6x - 2x + 8}{x(x + 2)} = \frac{4x + 8}{x(x + 2)} = \frac{4(x + 2)}{x(x + 2)} = \frac{4}{x}\]

\[y = \frac{4}{x};\ \ \ \ x \neq 0;\ \ x \neq - 2.\]

\[\frac{1}{3}\sqrt{144} + 5\sqrt{\frac{16}{225}} - \left( 0,2\sqrt{6} \right)^{2} =\]

\[= \frac{1}{3} \cdot 12 + 5 \cdot \frac{4}{15} - 0,04 \cdot 6 =\]

\[= 4 + \frac{4}{3} - 0,24 = 4 + \frac{4^{\backslash 25}}{3} - \frac{6^{\backslash 3}}{25} =\]

\[= 4 + \frac{100 - 18}{75} = 4 + \frac{82}{75} =\]

\[= 4 + 1\frac{7}{75} = 5\frac{7}{75}\]

\[\frac{\sqrt{98}}{\sqrt{2}} + \sqrt{150} \cdot \sqrt{6} - \sqrt{7^{4} \cdot 3^{2}} =\]

\[= \sqrt{\frac{98}{2}} + \sqrt{900} - 7^{2} \cdot 3 =\]

\[= \sqrt{49} + 30 - 49 \cdot 3 =\]

\[= 7 + 30 - 147 = - 110.\]

\[2\sqrt{x - 1} = 4\]

\[\sqrt{x - 1} = 2\]

\[\left( \sqrt{x - 1} \right)^{2} = 2^{2}\]

\[x - 1 = 4\]

\[x = 5.\]

\[3\sqrt{x + 2} > - 1\]

\[\sqrt{x + 2} > - \frac{1}{3}\]

\[x + 2 \geq 0\]

\[x \geq - 2.\]

\[\frac{1}{2}a^{4}\sqrt{36a^{6}}\ при\ a < 0\]

\[\frac{1}{2}a^{4}\sqrt{36a^{6}} = \frac{1}{2}a^{4}6\left| a^{3} \right| =\]

\[= 3a^{4} \cdot ( - a)^{3} = - 3a^{7}.\]

\[\frac{3x - 4}{\sqrt{x} - 3}\]

\[x \geq 0;\]

\[\sqrt{x} - 3 \neq 0\]

\[\sqrt{x} \neq 3\]

\[\left( \sqrt{x} \right)^{2} \neq 3^{2}\]

\[x \neq 9.\]

\[Ответ:x \geq 0;\ \ x \neq 9.\]

\[\frac{1}{7}\sqrt{196} + 3\sqrt{\frac{49}{324}} - \left( 0,3\sqrt{8} \right)^{2} =\]

\[= \frac{1}{7} \cdot 14 + 3 \cdot \frac{7}{18} - 0,09 \cdot 8 =\]

\[= 2 + \frac{7}{6} - 0,72 = 2 + \frac{7^{\backslash 25}}{6} - \frac{18^{\backslash 6}}{25} =\]

\[= 2 + \frac{175 - 108}{150} = 2 + \frac{67}{150} = 2\frac{67}{150}\]

\[\frac{\sqrt{128}}{\sqrt{2}} - \sqrt{75 \cdot 12} + \sqrt{5^{4} \cdot 3^{2}} =\]

\[= \sqrt{\frac{128}{2}} - \sqrt{25 \cdot 3 \cdot 3 \cdot 4} + 5^{2} \cdot 3 =\]

\[= \sqrt{64} - 3 \cdot 2 \cdot 5 + 25 \cdot 3 =\]

\[= 8 - 30 + 75 = 53\]

\[3\sqrt{x + 1} = 9\]

\[\sqrt{x + 1} = 3\]

\[\left( \sqrt{x + 1} \right)^{2} = 3^{2}\]

\[x + 1 = 9\]

\[x = 9 - 1\]

\[x = 8.\]