Страницы 44-45 номер 2.3.24, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Выполнить действия: 1) \(\frac{n-2}{n-5}:\left(\frac{n^2+24}{n^2-25}-\frac{4}{n-5}\right)\); 2) \(\left(\frac{m^2+2m-10}{(m-3)^2}-\frac{2}{m-3}\right):\frac{m-4}{m-3}\); 3) \(\frac{y-2x}{2x+y}+\frac{x^2+3x}{4x^2-y^2}:\frac{3+x}{4x+2y}\); 4) \(\frac{2-n}{m-3}+\frac{p^2-4}{m^2-9}:\frac{p^2-2p}{mn+3n}\); 5) \(\left(\frac{x}{x+y}+\frac{x^2}{y^2-x^2}\right):\left(\frac{x^2}{x+y}-\frac{x^3}{x^2+y^2+2xy}\right)\); 6) \(\left(\frac{1}{b-2a}+\frac{2a}{4a^2-b^2}\right):\left(\frac{2a}{b+2a}-\frac{4a^2}{4a^2+4ab+b^2}\right)\); 7) \(\frac{2m^3}{m^3+n^3}\cdot\frac{m+n}{m}-\frac{2m^2}{m^2-mn+n^2}\); 8) \(\frac{p+1}{p^3+p^2+p}:\frac{1}{p^4-p}-p^2\); 9) \(\frac{2+6a}{a}\cdot\left(\frac{1}{2-6a}+\frac{1}{27a^3-1}:\frac{1+3a}{1+3a+9a^2}\right)\); 10) \(\frac{8-b^3}{2+b}:\left(2+\frac{b^2}{2+b}\right)-\frac{b^2}{b-2}\cdot\frac{4-b^2}{b^2+2b}\); 11) \(\left(\frac{2m}{3+m}+\frac{9}{m^2-3m+9}-\frac{m^3-15m^2}{m^3+27}\right)\left(m+3-\frac{9m}{m+3}\right):(m+3)\); 12) \(\left(\frac{2n}{n+4}+\frac{16}{n^2-4n+16}-\frac{n^3-20n^2}{n^3+64}\right)\left(n+4-\frac{12n}{n+4}\right):(n+4)\).
1) \[\begin{aligned} &\frac{n^2 + 24}{n^2 - 25} - \frac{4}{n - 5} = \frac{n^2 + 24}{(n - 5)(n + 5)} - \frac{4}{n - 5} = {} \\ &= \frac{n^2 + 24 - 4(n + 5)}{(n - 5)(n + 5)} = \frac{n^2 - 4n + 4}{(n - 5)(n + 5)} = \frac{(n - 2)^2}{(n - 5)(n + 5)}; \end{aligned}\]
\[\frac{n - 2}{n - 5} : \frac{(n - 2)^2}{(n - 5)(n + 5)} = \frac{n - 2}{n - 5} \cdot \frac{(n - 5)(n + 5)}{(n - 2)^2} = \frac{n + 5}{n - 2}.\]
2) \[\begin{aligned} &\frac{m^2 + 2m - 10}{(m - 3)^2} - \frac{2}{m - 3} = \frac{m^2 + 2m - 10 - 2(m - 3)}{(m - 3)^2} = {} \\ &= \frac{m^2 + 2m - 10 - 2m + 6}{(m - 3)^2} = \frac{m^2 - 4}{(m - 3)^2}; \end{aligned}\]
\[\frac{m^2 - 4}{(m - 3)^2} : \frac{m - 4}{m - 3} = \frac{m^2 - 4}{(m - 3)^2} \cdot \frac{m - 3}{m - 4} = \frac{m^2 - 4}{(m - 3)(m - 4)}.\]
3) \[\frac{x^2 + 3x}{4x^2 - y^2} : \frac{3 + x}{4x + 2y} = \frac{x(x + 3)}{(2x - y)(2x + y)} \cdot \frac{2(2x + y)}{x + 3} = \frac{2x}{2x - y};\]
\[\begin{aligned} &\frac{y - 2x}{2x + y} + \frac{2x}{2x - y} = \frac{(y - 2x)(2x - y) + 2x(2x + y)}{(2x + y)(2x - y)} = {} \\ &= \frac{-4x^2 + 4xy - y^2 + 4x^2 + 2xy}{4x^2 - y^2} = \frac{6xy - y^2}{4x^2 - y^2}. \end{aligned}\]
4) \[\frac{p^2 - 4}{m^2 - 9} : \frac{p^2 - 2p}{mn + 3n} = \frac{(p - 2)(p + 2)}{(m - 3)(m + 3)} \cdot \frac{n(m + 3)}{p(p - 2)} = \frac{n(p + 2)}{p(m - 3)};\]
\[\frac{2 - n}{m - 3} + \frac{n(p + 2)}{p(m - 3)} = \frac{p(2 - n) + n(p + 2)}{p(m - 3)} = \frac{2p - np + np + 2n}{p(m - 3)} = \frac{2n + 2p}{p(m - 3)}.\]
5) \[\begin{aligned} &\frac{x}{x + y} + \frac{x^2}{y^2 - x^2} = \frac{x}{x + y} + \frac{x^2}{(y - x)(y + x)} = {} \\ &= \frac{x(y - x) + x^2}{(x + y)(y - x)} = \frac{xy - x^2 + x^2}{(x + y)(y - x)} = \frac{xy}{(x + y)(y - x)}; \end{aligned}\]
\[\begin{aligned} &\frac{x^2}{x + y} - \frac{x^3}{x^2 + y^2 + 2xy} = \frac{x^2}{x + y} - \frac{x^3}{(x + y)^2} = {} \\ &= \frac{x^2(x + y) - x^3}{(x + y)^2} = \frac{x^3 + x^2y - x^3}{(x + y)^2} = \frac{x^2y}{(x + y)^2}; \end{aligned}\]
\[\frac{xy}{(x + y)(y - x)} : \frac{x^2y}{(x + y)^2} = \frac{xy}{(x + y)(y - x)} \cdot \frac{(x + y)^2}{x^2y} = \frac{x + y}{x(y - x)}.\]
6) В первой дроби меняем знак знаменателя и знак перед дробью:
\[\begin{aligned} &\frac{1}{b - 2a} + \frac{2a}{4a^2 - b^2} = -\frac{1}{2a - b} + \frac{2a}{(2a - b)(2a + b)} = {} \\ &= \frac{-(2a + b) + 2a}{(2a - b)(2a + b)} = \frac{-b}{(2a - b)(2a + b)}; \end{aligned}\]
\[\begin{aligned} &\frac{2a}{b + 2a} - \frac{4a^2}{4a^2 + 4ab + b^2} = \frac{2a}{2a + b} - \frac{4a^2}{(2a + b)^2} = {} \\ &= \frac{2a(2a + b) - 4a^2}{(2a + b)^2} = \frac{4a^2 + 2ab - 4a^2}{(2a + b)^2} = \frac{2ab}{(2a + b)^2}; \end{aligned}\]
\[\begin{aligned} &\frac{-b}{(2a - b)(2a + b)} : \frac{2ab}{(2a + b)^2} = \frac{-b}{(2a - b)(2a + b)} \cdot \frac{(2a + b)^2}{2ab} = {} \\ &= \frac{-(2a + b)}{2a(2a - b)} = \frac{2a + b}{2a(b - 2a)}. \end{aligned}\]
7) \[\frac{2m^3}{m^3 + n^3} \cdot \frac{m + n}{m} = \frac{2m^3(m + n)}{(m + n)(m^2 - mn + n^2) \cdot m} = \frac{2m^2}{m^2 - mn + n^2};\]
\[\frac{2m^2}{m^2 - mn + n^2} - \frac{2m^2}{m^2 - mn + n^2} = 0.\]
8) \(p^3 + p^2 + p = p(p^2 + p + 1)\), \(p^4 - p = p(p^3 - 1) = p(p - 1)(p^2 + p + 1)\), поэтому
\[\begin{aligned} &\frac{p + 1}{p^3 + p^2 + p} : \frac{1}{p^4 - p} = \frac{p + 1}{p(p^2 + p + 1)} \cdot p(p - 1)(p^2 + p + 1) = {} \\ &= (p + 1)(p - 1) = p^2 - 1; \end{aligned}\]
\[p^2 - 1 - p^2 = -1.\]
9) \(27a^3 - 1 = (3a - 1)(9a^2 + 3a + 1)\), \(2 - 6a = -2(3a - 1)\), поэтому
\[\frac{1}{27a^3 - 1} : \frac{1 + 3a}{1 + 3a + 9a^2} = \frac{1}{(3a - 1)(9a^2 + 3a + 1)} \cdot \frac{9a^2 + 3a + 1}{1 + 3a} = \frac{1}{(3a - 1)(3a + 1)};\]
\[\begin{aligned} &\frac{1}{2 - 6a} + \frac{1}{(3a - 1)(3a + 1)} = -\frac{1}{2(3a - 1)} + \frac{1}{(3a - 1)(3a + 1)} = {} \\ &= \frac{-(3a + 1) + 2}{2(3a - 1)(3a + 1)} = \frac{-(3a - 1)}{2(3a - 1)(3a + 1)} = -\frac{1}{2(3a + 1)}; \end{aligned}\]
\[\frac{2 + 6a}{a} \cdot \left(-\frac{1}{2(3a + 1)}\right) = -\frac{2(1 + 3a)}{a \cdot 2(3a + 1)} = -\frac{1}{a}.\]
10) \[2 + \frac{b^2}{2 + b} = \frac{2(2 + b) + b^2}{2 + b} = \frac{b^2 + 2b + 4}{b + 2};\]
\[\frac{8 - b^3}{2 + b} : \frac{b^2 + 2b + 4}{b + 2} = \frac{(2 - b)(4 + 2b + b^2)}{b + 2} \cdot \frac{b + 2}{b^2 + 2b + 4} = 2 - b;\]
\[\begin{aligned} &\frac{b^2}{b - 2} \cdot \frac{4 - b^2}{b^2 + 2b} = \frac{b^2(2 - b)(2 + b)}{(b - 2) \cdot b(b + 2)} = {} \\ &= \frac{b(2 - b)}{b - 2} = \frac{-b(b - 2)}{b - 2} = -b; \end{aligned}\]
\[2 - b - (-b) = 2 - b + b = 2.\]
11) \(m^3 + 27 = (m + 3)(m^2 - 3m + 9)\), поэтому
\[\begin{aligned} &\frac{2m}{3 + m} + \frac{9}{m^2 - 3m + 9} - \frac{m^3 - 15m^2}{m^3 + 27} = {} \\ &= \frac{2m(m^2 - 3m + 9) + 9(m + 3) - (m^3 - 15m^2)}{(m + 3)(m^2 - 3m + 9)} = {} \\ &= \frac{2m^3 - 6m^2 + 18m + 9m + 27 - m^3 + 15m^2}{(m + 3)(m^2 - 3m + 9)} = {} \\ &= \frac{m^3 + 9m^2 + 27m + 27}{(m + 3)(m^2 - 3m + 9)} = \frac{(m + 3)^3}{(m + 3)(m^2 - 3m + 9)} = \frac{(m + 3)^2}{m^2 - 3m + 9}; \end{aligned}\]
\[m + 3 - \frac{9m}{m + 3} = \frac{(m + 3)^2 - 9m}{m + 3} = \frac{m^2 + 6m + 9 - 9m}{m + 3} = \frac{m^2 - 3m + 9}{m + 3};\]
\[\frac{(m + 3)^2}{m^2 - 3m + 9} \cdot \frac{m^2 - 3m + 9}{m + 3} = m + 3;\]
\[(m + 3) : (m + 3) = 1.\]
12) \(n^3 + 64 = (n + 4)(n^2 - 4n + 16)\), поэтому
\[\begin{aligned} &\frac{2n}{n + 4} + \frac{16}{n^2 - 4n + 16} - \frac{n^3 - 20n^2}{n^3 + 64} = {} \\ &= \frac{2n(n^2 - 4n + 16) + 16(n + 4) - (n^3 - 20n^2)}{(n + 4)(n^2 - 4n + 16)} = {} \\ &= \frac{2n^3 - 8n^2 + 32n + 16n + 64 - n^3 + 20n^2}{(n + 4)(n^2 - 4n + 16)} = {} \\ &= \frac{n^3 + 12n^2 + 48n + 64}{(n + 4)(n^2 - 4n + 16)} = \frac{(n + 4)^3}{(n + 4)(n^2 - 4n + 16)} = \frac{(n + 4)^2}{n^2 - 4n + 16}; \end{aligned}\]
\[n + 4 - \frac{12n}{n + 4} = \frac{(n + 4)^2 - 12n}{n + 4} = \frac{n^2 + 8n + 16 - 12n}{n + 4} = \frac{n^2 - 4n + 16}{n + 4};\]
\[\frac{(n + 4)^2}{n^2 - 4n + 16} \cdot \frac{n^2 - 4n + 16}{n + 4} = n + 4;\]
\[(n + 4) : (n + 4) = 1.\]
Ответ: 1) \(\frac{n + 5}{n - 2}\); 2) \(\frac{m^2 - 4}{(m - 3)(m - 4)}\); 3) \(\frac{6xy - y^2}{4x^2 - y^2}\); 4) \(\frac{2n + 2p}{p(m - 3)}\); 5) \(\frac{x + y}{x(y - x)}\); 6) \(\frac{2a + b}{2a(b - 2a)}\); 7) \(0\); 8) \(-1\); 9) \(-\frac{1}{a}\); 10) \(2\); 11) \(1\); 12) \(1\).
Выполнить действия: 1) \(\frac{n-2}{n-5}:\left(\frac{n^2+24}{n^2-25}-\frac{4}{n-5}\right)\); 2) \(\left(\frac{m^2+2m-10}{(m-3)^2}-\frac{2}{m-3}\right):\frac{m-4}{m-3}\); 3) \(\frac{y-2x}{2x+y}+\frac{x^2+3x}{4x^2-y^2}:\frac{3+x}{4x+2y}\); 4) \(\frac{2-n}{m-3}+\frac{p^2-4}{m^2-9}:\frac{p^2-2p}{mn+3n}\); 5) \(\left(\frac{x}{x+y}+\frac{x^2}{y^2-x^2}\right):\left(\frac{x^2}{x+y}-\frac{x^3}{x^2+y^2+2xy}\right)\); 6) \(\left(\frac{1}{b-2a}+\frac{2a}{4a^2-b^2}\right):\left(\frac{2a}{b+2a}-\frac{4a^2}{4a^2+4ab+b^2}\right)\); 7) \(\frac{2m^3}{m^3+n^3}\cdot\frac{m+n}{m}-\frac{2m^2}{m^2-mn+n^2}\); 8) \(\frac{p+1}{p^3+p^2+p}:\frac{1}{p^4-p}-p^2\); 9) \(\frac{2+6a}{a}\cdot\left(\frac{1}{2-6a}+\frac{1}{27a^3-1}:\frac{1+3a}{1+3a+9a^2}\right)\); 10) \(\frac{8-b^3}{2+b}:\left(2+\frac{b^2}{2+b}\right)-\frac{b^2}{b-2}\cdot\frac{4-b^2}{b^2+2b}\); 11) \(\left(\frac{2m}{3+m}+\frac{9}{m^2-3m+9}-\frac{m^3-15m^2}{m^3+27}\right)\left(m+3-\frac{9m}{m+3}\right):(m+3)\); 12) \(\left(\frac{2n}{n+4}+\frac{16}{n^2-4n+16}-\frac{n^3-20n^2}{n^3+64}\right)\left(n+4-\frac{12n}{n+4}\right):(n+4)\).
Перед сложением и вычитанием знаменатели раскладываем на множители и приводим дроби к общему знаменателю, деление заменяем умножением на дробь, обратную делителю, а сокращаем дробь только после разложения её числителя и знаменателя на множители.
1) \[\begin{aligned} &\frac{n^2 + 24}{n^2 - 25} - \frac{4}{n - 5} = \frac{n^2 + 24}{(n - 5)(n + 5)} - \frac{4}{n - 5} = {} \\ &= \frac{n^2 + 24 - 4(n + 5)}{(n - 5)(n + 5)} = \frac{n^2 - 4n + 4}{(n - 5)(n + 5)} = \frac{(n - 2)^2}{(n - 5)(n + 5)}; \end{aligned}\]
\[\frac{n - 2}{n - 5} : \frac{(n - 2)^2}{(n - 5)(n + 5)} = \frac{n - 2}{n - 5} \cdot \frac{(n - 5)(n + 5)}{(n - 2)^2} = \frac{n + 5}{n - 2}.\]
2) \[\begin{aligned} &\frac{m^2 + 2m - 10}{(m - 3)^2} - \frac{2}{m - 3} = \frac{m^2 + 2m - 10 - 2(m - 3)}{(m - 3)^2} = {} \\ &= \frac{m^2 + 2m - 10 - 2m + 6}{(m - 3)^2} = \frac{m^2 - 4}{(m - 3)^2}; \end{aligned}\]
\[\frac{m^2 - 4}{(m - 3)^2} : \frac{m - 4}{m - 3} = \frac{m^2 - 4}{(m - 3)^2} \cdot \frac{m - 3}{m - 4} = \frac{m^2 - 4}{(m - 3)(m - 4)}.\]
3) \[\frac{x^2 + 3x}{4x^2 - y^2} : \frac{3 + x}{4x + 2y} = \frac{x(x + 3)}{(2x - y)(2x + y)} \cdot \frac{2(2x + y)}{x + 3} = \frac{2x}{2x - y};\]
\[\begin{aligned} &\frac{y - 2x}{2x + y} + \frac{2x}{2x - y} = \frac{(y - 2x)(2x - y) + 2x(2x + y)}{(2x + y)(2x - y)} = {} \\ &= \frac{-4x^2 + 4xy - y^2 + 4x^2 + 2xy}{4x^2 - y^2} = \frac{6xy - y^2}{4x^2 - y^2}. \end{aligned}\]
4) \[\frac{p^2 - 4}{m^2 - 9} : \frac{p^2 - 2p}{mn + 3n} = \frac{(p - 2)(p + 2)}{(m - 3)(m + 3)} \cdot \frac{n(m + 3)}{p(p - 2)} = \frac{n(p + 2)}{p(m - 3)};\]
\[\frac{2 - n}{m - 3} + \frac{n(p + 2)}{p(m - 3)} = \frac{p(2 - n) + n(p + 2)}{p(m - 3)} = \frac{2p - np + np + 2n}{p(m - 3)} = \frac{2n + 2p}{p(m - 3)}.\]
5) \[\begin{aligned} &\frac{x}{x + y} + \frac{x^2}{y^2 - x^2} = \frac{x}{x + y} + \frac{x^2}{(y - x)(y + x)} = {} \\ &= \frac{x(y - x) + x^2}{(x + y)(y - x)} = \frac{xy - x^2 + x^2}{(x + y)(y - x)} = \frac{xy}{(x + y)(y - x)}; \end{aligned}\]
\[\begin{aligned} &\frac{x^2}{x + y} - \frac{x^3}{x^2 + y^2 + 2xy} = \frac{x^2}{x + y} - \frac{x^3}{(x + y)^2} = {} \\ &= \frac{x^2(x + y) - x^3}{(x + y)^2} = \frac{x^3 + x^2y - x^3}{(x + y)^2} = \frac{x^2y}{(x + y)^2}; \end{aligned}\]
\[\frac{xy}{(x + y)(y - x)} : \frac{x^2y}{(x + y)^2} = \frac{xy}{(x + y)(y - x)} \cdot \frac{(x + y)^2}{x^2y} = \frac{x + y}{x(y - x)}.\]
6) В первой дроби меняем знак знаменателя и знак перед дробью:
\[\begin{aligned} &\frac{1}{b - 2a} + \frac{2a}{4a^2 - b^2} = -\frac{1}{2a - b} + \frac{2a}{(2a - b)(2a + b)} = {} \\ &= \frac{-(2a + b) + 2a}{(2a - b)(2a + b)} = \frac{-b}{(2a - b)(2a + b)}; \end{aligned}\]
\[\begin{aligned} &\frac{2a}{b + 2a} - \frac{4a^2}{4a^2 + 4ab + b^2} = \frac{2a}{2a + b} - \frac{4a^2}{(2a + b)^2} = {} \\ &= \frac{2a(2a + b) - 4a^2}{(2a + b)^2} = \frac{4a^2 + 2ab - 4a^2}{(2a + b)^2} = \frac{2ab}{(2a + b)^2}; \end{aligned}\]
\[\begin{aligned} &\frac{-b}{(2a - b)(2a + b)} : \frac{2ab}{(2a + b)^2} = \frac{-b}{(2a - b)(2a + b)} \cdot \frac{(2a + b)^2}{2ab} = {} \\ &= \frac{-(2a + b)}{2a(2a - b)} = \frac{2a + b}{2a(b - 2a)}. \end{aligned}\]
7) \[\frac{2m^3}{m^3 + n^3} \cdot \frac{m + n}{m} = \frac{2m^3(m + n)}{(m + n)(m^2 - mn + n^2) \cdot m} = \frac{2m^2}{m^2 - mn + n^2};\]
\[\frac{2m^2}{m^2 - mn + n^2} - \frac{2m^2}{m^2 - mn + n^2} = 0.\]
8) \(p^3 + p^2 + p = p(p^2 + p + 1)\), \(p^4 - p = p(p^3 - 1) = p(p - 1)(p^2 + p + 1)\), поэтому
\[\begin{aligned} &\frac{p + 1}{p^3 + p^2 + p} : \frac{1}{p^4 - p} = \frac{p + 1}{p(p^2 + p + 1)} \cdot p(p - 1)(p^2 + p + 1) = {} \\ &= (p + 1)(p - 1) = p^2 - 1; \end{aligned}\]
\[p^2 - 1 - p^2 = -1.\]
9) \(27a^3 - 1 = (3a - 1)(9a^2 + 3a + 1)\), \(2 - 6a = -2(3a - 1)\), поэтому
\[\frac{1}{27a^3 - 1} : \frac{1 + 3a}{1 + 3a + 9a^2} = \frac{1}{(3a - 1)(9a^2 + 3a + 1)} \cdot \frac{9a^2 + 3a + 1}{1 + 3a} = \frac{1}{(3a - 1)(3a + 1)};\]
\[\begin{aligned} &\frac{1}{2 - 6a} + \frac{1}{(3a - 1)(3a + 1)} = -\frac{1}{2(3a - 1)} + \frac{1}{(3a - 1)(3a + 1)} = {} \\ &= \frac{-(3a + 1) + 2}{2(3a - 1)(3a + 1)} = \frac{-(3a - 1)}{2(3a - 1)(3a + 1)} = -\frac{1}{2(3a + 1)}; \end{aligned}\]
\[\frac{2 + 6a}{a} \cdot \left(-\frac{1}{2(3a + 1)}\right) = -\frac{2(1 + 3a)}{a \cdot 2(3a + 1)} = -\frac{1}{a}.\]
10) \[2 + \frac{b^2}{2 + b} = \frac{2(2 + b) + b^2}{2 + b} = \frac{b^2 + 2b + 4}{b + 2};\]
\[\frac{8 - b^3}{2 + b} : \frac{b^2 + 2b + 4}{b + 2} = \frac{(2 - b)(4 + 2b + b^2)}{b + 2} \cdot \frac{b + 2}{b^2 + 2b + 4} = 2 - b;\]
\[\begin{aligned} &\frac{b^2}{b - 2} \cdot \frac{4 - b^2}{b^2 + 2b} = \frac{b^2(2 - b)(2 + b)}{(b - 2) \cdot b(b + 2)} = {} \\ &= \frac{b(2 - b)}{b - 2} = \frac{-b(b - 2)}{b - 2} = -b; \end{aligned}\]
\[2 - b - (-b) = 2 - b + b = 2.\]
11) \(m^3 + 27 = (m + 3)(m^2 - 3m + 9)\), поэтому
\[\begin{aligned} &\frac{2m}{3 + m} + \frac{9}{m^2 - 3m + 9} - \frac{m^3 - 15m^2}{m^3 + 27} = {} \\ &= \frac{2m(m^2 - 3m + 9) + 9(m + 3) - (m^3 - 15m^2)}{(m + 3)(m^2 - 3m + 9)} = {} \\ &= \frac{2m^3 - 6m^2 + 18m + 9m + 27 - m^3 + 15m^2}{(m + 3)(m^2 - 3m + 9)} = {} \\ &= \frac{m^3 + 9m^2 + 27m + 27}{(m + 3)(m^2 - 3m + 9)} = \frac{(m + 3)^3}{(m + 3)(m^2 - 3m + 9)} = \frac{(m + 3)^2}{m^2 - 3m + 9}; \end{aligned}\]
\[m + 3 - \frac{9m}{m + 3} = \frac{(m + 3)^2 - 9m}{m + 3} = \frac{m^2 + 6m + 9 - 9m}{m + 3} = \frac{m^2 - 3m + 9}{m + 3};\]
\[\frac{(m + 3)^2}{m^2 - 3m + 9} \cdot \frac{m^2 - 3m + 9}{m + 3} = m + 3;\]
\[(m + 3) : (m + 3) = 1.\]
12) \(n^3 + 64 = (n + 4)(n^2 - 4n + 16)\), поэтому
\[\begin{aligned} &\frac{2n}{n + 4} + \frac{16}{n^2 - 4n + 16} - \frac{n^3 - 20n^2}{n^3 + 64} = {} \\ &= \frac{2n(n^2 - 4n + 16) + 16(n + 4) - (n^3 - 20n^2)}{(n + 4)(n^2 - 4n + 16)} = {} \\ &= \frac{2n^3 - 8n^2 + 32n + 16n + 64 - n^3 + 20n^2}{(n + 4)(n^2 - 4n + 16)} = {} \\ &= \frac{n^3 + 12n^2 + 48n + 64}{(n + 4)(n^2 - 4n + 16)} = \frac{(n + 4)^3}{(n + 4)(n^2 - 4n + 16)} = \frac{(n + 4)^2}{n^2 - 4n + 16}; \end{aligned}\]
\[n + 4 - \frac{12n}{n + 4} = \frac{(n + 4)^2 - 12n}{n + 4} = \frac{n^2 + 8n + 16 - 12n}{n + 4} = \frac{n^2 - 4n + 16}{n + 4};\]
\[\frac{(n + 4)^2}{n^2 - 4n + 16} \cdot \frac{n^2 - 4n + 16}{n + 4} = n + 4;\]
\[(n + 4) : (n + 4) = 1.\]
Ответ: 1) \(\frac{n + 5}{n - 2}\); 2) \(\frac{m^2 - 4}{(m - 3)(m - 4)}\); 3) \(\frac{6xy - y^2}{4x^2 - y^2}\); 4) \(\frac{2n + 2p}{p(m - 3)}\); 5) \(\frac{x + y}{x(y - x)}\); 6) \(\frac{2a + b}{2a(b - 2a)}\); 7) \(0\); 8) \(-1\); 9) \(-\frac{1}{a}\); 10) \(2\); 11) \(1\); 12) \(1\).
