Предмет: Математика, автор: nazarvovchenkooo

сделайте то что позначено галочками

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Ответы

Автор ответа: Ляляляля109

16.1

$\frac{x}{6} + \frac{y}{6} = \frac{x + y}{6}$

$\frac{a}{3} - \frac{b}{3} = \frac{a - b}{3}$

$\frac{m}{n} + \frac{4m}{n} = \frac{m + 4m}{n} = \frac{5m}{n}$

$\frac{m + n}{6} - \frac{m - 2n}{6} = \frac{m + n - m + 2n}{6} = \frac{3n}{6} = \frac{n}{2}$

$\frac{2a - 3b}{6ab} + \frac{9b - 2a}{6ab} = \frac{2a - 3b + 9b - 2a}{6ab} = \frac{6b}{6ab} = \frac{1}{a}$

$\frac{8m + 3}{10 {m}^{2} } - \frac{2m + 3}{10 {m}^{2} } = \frac{8m+3 - 2m - 3 }{10 {m}^{2} } = \frac{6m}{10 {m}^{2} } = \frac{3}{5m}$

16.2

$\frac{a - b}{2b} - \frac{a}{2b} = \frac{a - b - a}{2b} = - \frac{b}{2b} = - \frac{1}{2}$

$- \frac{a - 12b}{27a} + \frac{a + 15b}{27a} = \frac{a + 15b - a + 12b}{27a} = \frac{27b}{27a} = \frac{b}{a}$

$\frac{10a + 6b}{11 {a}^{3} } - \frac{6b - a}{11 {a}^{3} } = \frac{10a + 6b - 6b + a}{11 {a}^{3} } = \frac{11a}{11 {a}^{3} } = \frac{1}{ {a}^{2} }$

$\frac{ {x}^{2} - xy }{ {x}^{2}y } + \frac{2xy - 3 {x}^{2} }{ {x}^{2} y} = \frac{ {x}^{2} - xy + 2xy - 3 {x}^{2} }{ {x}^{2} y} = \frac{xy - 2 {x}^{2} }{ {x}^{2}y } = \frac{x(y - 2x)}{ {x}^{2} y} = \frac{y - 2x}{xy}$

16.3

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

$\frac{t}{ {t}^{2} - 16 } - \frac{4}{ {t}^{2} - 16 } = \frac{t - 4}{(t - 4)(t + 4)} = \frac{1}{t + 4}$

$\frac{ {m}^{2} }{ {(m - 5)}^{2} } - \frac{25}{ {(m - 5)}^{2} } = \frac{ {m}^{2} - 25}{ {(m - 5)}^{2} } = \frac{(m - 5)(m + 5)}{ {(m - 5)}^{2} } = \frac{m + 5}{m - 5}$