Question

помогите пожалуйста с алгеброй даю 40 баллов​

Answer 1

1 задание

2) а)

$\frac{ {(a - b)}^{2} }{18b} - \frac{ {(a - b)}^{2} }{12b} + \frac{ {a}^{2} - {b}^{2} }{36b} = \frac{ {(a - b)}^{2} \times 2 }{18b \times 2} - \frac{ {(a - b)}^{2} \times 3}{12b \times 3} + \frac{ {a}^{2} - {b}^{2} }{36b} = \frac{2 \times {(a - b)}^{2} - 3 \times {(a - b)}^{2} + {a}^{2} - {b}^{2} }{36b} = \frac{ - {(a - b)}^{2} + {a}^{2} - {b}^{2} }{36b} = \frac{ - {a}^{2} + 2ab - {b}^{2} + {a}^{2} - {b}^{2} }{36b} = \frac{2ab - {b}^{2} }{36b} = \frac{b(2a - b)}{36b} = \frac{2a - b}{36}$

б)

$\frac{3x + 2}{5x} - \frac{5x + 3y}{10xy} - \frac{y - 1}{2y} = \frac{(3x + 2) \times 2y}{5x \times 2y} - \frac{5x + 3y}{10xy} - \frac{(y - 1) \times 5x}{2y \times 5x} = \frac{6xy + 4y}{10xy} - \frac{5x + 3y}{10xy} - \frac{5xy - 5x}{10xy} = \frac{6xy + 4y - 5x - 3y - 5xy + 5x}{10xy} = \frac{xy + y}{10xy} = \frac{y(x + 1)}{10xy} = \frac{x + 1}{10x}$

3) а)

$\frac{c - 2}{3(c + 4)} + \frac{c}{c + 4} = \frac{c - 2}{3(c + 4)} + \frac{c \times 3}{(c + 4) \times 3} = \frac{c - 2}{3c + 12} + \frac{3c}{3c + 12} = \frac{c - 2 + 3c}{3c + 12} = \frac{4c - 2}{3c + 12}$

б)

$\frac{b - 2}{2b - 6} - \frac{b - 1}{3b - 3} = \frac{b - 2}{2(b - 2)} - \frac{b - 1}{3(b - 1)} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$

в)

$\frac{4a}{3a - 6} + \frac{3a}{8 - 4a} = \frac{4a}{3(a - 2)} + \frac{3a}{ - 4(a - 2)} = \frac{4a \times ( - 4)}{3(a - 2) \times ( - 4)} + \frac{3a \times 3}{ - 4(a - 2) \times 3} = \frac{ - 16a}{ - 12(a - 2)} + \frac{9a}{ - 12(a - 2)} = \frac{ - 16a + 9a}{ - 12(a - 2)} = \frac{ - 7a}{ - (12a - 24)} = \frac{7a}{12a - 24}$

4) а)

$\frac{x + 4}{xy - {x}^{2} } + \frac{y + 4}{xy - {y}^{2} } = \frac{x + 4}{x(y - x)} + \frac{y + 4}{ - y(y - x)} = \frac{(x + 4) \times ( - y)}{ x(y - x) \times ( - y)} + \frac{(y + 4) \times x}{ - y(y - x) \times x} = \frac{ - xy - 4y}{ - xy(y - x)} + \frac{xy + 4x}{ - xy(y - x)} = \frac{ - xy - 4y + xy + 4x}{ - xy(y - x)} = \frac{ - 4y + 4x}{ - xy(y - x)} = \frac{ - 4(y - x)}{ - xy(y - x)} = \frac{4}{xy}$

б)

$\frac{3a(x - 9a)}{ {x}^{2} - 3ax } - \frac{3 {a}^{2} - {x}^{2} }{ax - 3 {a}^{2} } = \frac{3ax - 27 {a}^{2} }{x(x - 3a)} - \frac{3 {a}^{2} - {x}^{2} }{a(x - 3a)} = \frac{(3ax - 27 {a}^{2} )\times a }{x(x - 3a) \times a} - \frac{(3 {a}^{2} - {x}^{2} ) \times x }{a(x - 3a) \times x} = \frac{3 {a}^{2} x - 27 {a}^{3} - 3 {a}^{2} x + {x}^{3} }{ax(x - 3a)} = \frac{ - 27 {a}^{3} + {x}^{3} }{ax(x - 3a)} = \frac{ {x}^{3} - {(3a)}^{3} }{ax(x - 3a)} = \frac{(x - 3a)( {x}^{2} + 3ax + 9 {a}^{2}) }{ax(x - 3a)} = \frac{ {x}^{2} + 3ax + 9 {a}^{2} }{ax}$

в)

$\frac{4}{ {c}^{2} - 9} - \frac{2}{ {c}^{2} + 3c } = \frac{4}{(c - 3)(c + 3)} - \frac{2}{c(c + 3)} = \frac{4 \times c}{(c - 3)(c + 3) \times c} - \frac{2 \times (c - 3)}{c(c + 3) \times (c - 3)} = \frac{4c}{c(c - 3)(c + 3)} - \frac{2c - 6}{c(c - 3)(c + 3) } = \frac{4c - 2c + 6}{c(c - 3)(c + 3)} = \frac{2c + 6}{c(c - 3)(c + 3)} = \frac{2(c + 3)}{c(c - 3)(c + 3)} = \frac{2}{c(c - 3)} = \frac{2}{ {c}^{2} - 3c}$

2 задание

2) а)

$\frac{6 {c}^{2} }{3c - 2} - 2c - 5 = \frac{6 {c}^{2} }{3c - 2} - \frac{2c \times (3c - 2)}{3c - 2} - \frac{5 \times (3c - 2)}{3c - 2} = \frac{6 {c}^{2} }{3c - 2} - \frac{6 {c}^{2} - 4c}{3c - 2} - \frac{15c - 10}{3c - 2} = \frac{6 {c}^{2} - 6 {c}^{2} + 4c - 15c + 10}{3c - 2} = \frac{ - 11c + 10}{3c - 2}$

б)

$2y - \frac{2 - 5y + 3 {y}^{2} }{3y - 2} - 1 = \frac{2y \times (3y - 2)}{3y - 2} - \frac{2 - 5y + 3 {y}^{2} }{3y - 2} - \frac{3y - 2}{3y - 2} = \frac{6 {y}^{2} - 4y }{3y - 2} - \frac{2 - 5y + 3 {y}^{2} }{3y - 2} - \frac{3y - 2}{3y - 2} = \frac{6 {y}^{2} - 4y - 2 + 5y - 3 {y}^{2} - 3y + 2 }{3y - 2} = \frac{3 {y}^{2} - 2y }{3 y - 2 } = \frac{y(3y - 2)}{3y - 2} = \frac{y}{1} = y$