Question

, как делать первый номер и последний, я хз как их делать

Answer 1

Номер 1.

1)

$\frac{ {m}^{6} {n}^{7} + {m}^{4} {n}^{9} }{ {m}^{4} {n}^{7} } = \frac{ {m}^{4} {n}^{7} ( {m}^{2} + {n}^{2} ) }{ {m}^{4} {n}^{7} } = {m}^{2} + {n}^{2}$

$m = - 0.9, \: n = - 0.1$

${m}^{2} + {n}^{2} = {( - 0.9)}^{2} + {( - 0.1)}^{2} = 0.81 + 0.01 = 0.82$

2)

$\frac{3 {a}^{3} - 48a }{5 {a}^{3} - 40 {a}^{2} + 80a } = \frac{3a( {a}^{2} - 16)}{5a( {a}^{2} - 8a + 16)} = \frac{3( {a}^{2} - {4}^{2}) }{5( {a - 4)}^{2} } = \frac{3(a - 4)(a + 4)}{5(a - 4)(a - 4)} = \frac{3(a + 4)}{5(a - 4)} = \frac{3a + 12}{5a - 20}$

$a = 16$

$\frac{3a + 12}{5a - 20} = \frac{3 \times 16 + 12}{5 \times 16 - 20} = \frac{60}{60} = 1$

3)

$\frac{ {(5a + 5b)}^{2} }{5 {a}^{2} - 5 {b}^{2} } = \frac{(5a + 5b)(5a + 5b)}{5( {a}^{2} - {b}^{2} )} = \frac{5(a + b)(5a + 5b)}{5(a - b)(a + b)} = \frac{5a + 5b}{a - b}$

$a = 0.3, b = - 0.2$

$\frac{5a + 5b}{a - b} = \frac{5 \times 0.3 + 5 \times ( - 0.2) }{0.3 - ( - 0.2)} = \frac{1.5 - 1}{0.5} = \frac{0.5}{0.5} = 1$

4)

$\frac{20 {x}^{2} - 60xy + 45 {y}^{2} }{21y - 14x} = \frac{5(4 {x}^{2} - 12xy + 9 {y}^{2} ) }{ - 7(2x - 3y)} = \frac{5( {2x - 3y)}^{2} }{ - 7(2x - 3y)} = - \frac{5(2x - 3y)}{7} = - \frac{5}{7} (2x - 3y)$

$2x - 3y = 0.7$

$- \frac{5}{7} (2x - 3y) = - \frac{5}{7} \times 0.7 = - \frac{5}{7} \times \frac{7}{10} = - \frac{1}{2} = - 0.5$

Номер 3.

1)

$\frac{ - {x}^{2} + 5x}{1 - 6x} + \frac{41 {x}^{2} - 2x }{6x - 1} = - \frac{ - {x}^{2} + 5x}{6x - 1} + \frac{41 {x}^{2} - 2x}{6x - 1} = \frac{ {x}^{2} - 5x + 41 {x}^{2} - 2x}{6x - 1} = \frac{42 {x}^{2} - 7x}{6x - 1} = \frac{7x(6x - 1)}{6x - 1} = 7x$

$x = \frac{1}{28}$

$7x = 7 \times \frac{1}{28} = \frac{7}{28} = \frac{1}{4} = 0.25$

2)

$\frac{ {(m - 1)}^{2} }{ {m}^{3} + 27 } + \frac{8 - m}{ {m}^{3} + 27} = \frac{ {m}^{2} - 2m + 1 + 8 - m }{ {m}^{3} + 27} = \frac{ {m}^{2} - 3m + 9}{(m + 3)( {m}^{2} - 3m + 9)} = \frac{1}{m + 3}$

$m = - 3.5$

$\frac{1}{m + 3} = \frac{1}{ - 3.5 + 3} = \frac{1}{ - 0.5} = - \frac{1}{ \frac{1}{2} } = - 1 \times 2 = - 2$

3)

$\frac{4 {c}^{2} - 8c}{3c - 2} - \frac{2c + 5 {c}^{2} }{2 - 3c} = \frac{4 {c}^{2} - 8c }{3c - 2} + \frac{2c + 5 {c}^{2} }{3c - 2} = \frac{4 {c}^{2} - 8c + 2c + 5 {c}^{2} }{3c - 2} = \frac{9 {c}^{2} - 6c}{3c - 2} = \frac{3c(3c - 2)}{3c - 2} = 3c$

$c = \frac{2}{9}$

$3c = 3 \times \frac{2}{9} = \frac{3 \times 2}{9} = \frac{6}{9} = \frac{2}{3}$

4)

$\frac{ {n}^{2} + n + 1 }{ {n}^{3} - 8 } - \frac{n + 3}{8 - {n}^{3} } = \frac{ {n}^{2} + n + 1}{ {n}^{3} - 8} + \frac{n + 3}{ {n}^{3} - 8 } = \frac{ {n}^{2} + n + 1 + n + 3}{ {n}^{3} - 8} = \frac{ {n}^{2} + 2n + 4 }{(n - 2)( {n}^{2} + 2n + 4)} = \frac{1}{n - 2}$

$n = - 4$

$\frac{1}{n - 2} = \frac{1}{ - 4 - 2} = \frac{1}{ - 6} = - \frac{1}{6}$