Задание 515

Преобразуйте в алгебраическую дробь:
а) $\frac{1}{a} + \frac{1}{a + b}$;
б) $\frac{1}{a + b} + \frac{1}{a - b}$;
в) $\frac{1}{m + n} - \frac{1}{n}$;
г) $\frac{1}{x - y} - \frac{1}{x + y}$;
д) $\frac{2}{a - b} + \frac{3}{a + b}$;
е) $\frac{4}{p - q} - \frac{3}{p + q}$;
ж) $\frac{2a}{a - 2b} + \frac{3a}{a + b}$;
з) $\frac{3x}{x - y} - \frac{2x}{2x - y}$;
и) $\frac{5m}{2m - n} - \frac{3m}{n - m}$;
к) $\frac{4p}{q - 2p} - \frac{2p}{2p + q}$;
л) $\frac{7}{2x - y} - \frac{5}{y - 2x}$;
м) $\frac{5x}{x - 3y} + \frac{4x + 3y}{3y - x}$.

Решение

а) $\frac{1}{a} + \frac{1}{a + b} = \frac{a + b + a}{a(a + b)} = \frac{2a + b}{a(a + b)}$

б) $\frac{1}{a + b} + \frac{1}{a - b} = \frac{a - b + a + b}{(a - b)(a + b)} = \frac{2a}{a^2 - b^2}$

в) $\frac{1}{m + n} - \frac{1}{n} = \frac{n - (m + n)}{n(m + n)} = \frac{n - m - n}{n(m + n)} = \frac{-m}{n(m + n)} = -\frac{m}{n(m + n)}$

г) $\frac{1}{x - y} - \frac{1}{x + y} = \frac{x + y - (x - y)}{(x - y)(x + y)} = \frac{x + y - x + y}{x^2 - y^2} = \frac{2y}{x^2 - y^2}$

д) $\frac{2}{a - b} + \frac{3}{a + b} = \frac{2(a + b) + 3(a - b)}{a^2 - b^2} = \frac{2a + 2b + 3a - 3b}{(a - b)(a + b)} = \frac{5a - b}{a^2 - b^2}$

е) $\frac{4}{p - q} - \frac{3}{p + q} = \frac{4(p + q) - 3(p - q)}{p^2 - q^2} = \frac{4p + 4q - 3p + 3q}{p^2 - q^2} = \frac{p + 7q}{p^2 - q^2}$

ж) $\frac{2a}{a - 2b} + \frac{3a}{a + b} = \frac{2a(a + b) + 3a(a - 2b)}{(a - 2b)(a + b)} = \frac{2a^2 + 2ab + 3a^2 - 6ab}{(a - 2b)(a + b)} = \frac{5a^2 - 4ab}{(a - 2b)(a + b)}$

з) $\frac{3x}{x - y} - \frac{2x}{2x - y} = \frac{3x(2x - y) - 2x(x - y)}{(x - y)(2x - y)} = \frac{6x^2 - 3xy - 2x^2 + 2xy)}{(x - y)(2x - y)} = \frac{4x^2 - xy}{(x - y)(2x - y)}$

и) $\frac{5m}{2m - n} - \frac{3m}{n - m} = \frac{5m(n - m) - 3m(2m - n)}{(2m - n)(n - m)} = \frac{5mn - 5m^2 - 6m^2 + 3mn}{(2m - n)(n - m)} = \frac{-11m^2 + 8mn}{(2m - n)(n - m)}$

к) $\frac{4p}{q - 2p} - \frac{2p}{2p + q} = \frac{4p(2p + q) - 2p(q - 2p)}{(q - 2p)(q + 2p)} = \frac{8p^2 + 4pq - 2pq + 4p^2}{q^2 - 4p^2} = \frac{12p^2 + 2pq}{q^2 - 4p^2}$

л) $\frac{7}{2x - y} - \frac{5}{y - 2x} = \frac{7}{2x - y} + \frac{5}{2x - y} = \frac{7 + 5}{2x - y} = \frac{12}{2x - y}$

м) $\frac{5x}{x - 3y} + \frac{4x + 3y}{3y - x} = \frac{5x}{x - 3y} - \frac{4x + 3y}{x - 3y} = \frac{5x - (4x + 3y)}{x - 3y} = \frac{5x - 4x - 3y}{x - 3y} = \frac{x - 3y}{x - 3y} = 1$

Задание 516

Преобразуйте в алгебраическую дробь:
а) $\frac{x}{8} - \frac{x}{4}$;
б) $\frac{a}{6} + \frac{a}{8}$;
в) $\frac{m^2}{3} - \frac{2m}{2}$;
г) $\frac{a - 1}{10} + \frac{a}{15}$;
д) $\frac{2x + 3}{6} + \frac{x - 1}{8}$;
е) $\frac{a - 3}{10} - \frac{2 - a}{15}$.

Решение

а) $\frac{x}{8} - \frac{x}{4} = \frac{x - 2x}{8} = \frac{-x}{8} = -\frac{x}{8}$

б) $\frac{a}{6} + \frac{a}{8} = \frac{4a + 3a}{24} = \frac{7a}{24}$

в) $\frac{m^2}{3} - \frac{2m}{2} = \frac{2m^2 - 6m}{6} = \frac{2(m^2 - 3m)}{6} = \frac{2m(m - 3)}{6} = \frac{m(m - 3)}{3}$

г) $\frac{a - 1}{10} + \frac{a}{15} = \frac{3(a - 1) + 2a}{30} = \frac{3a - 3 + 2a}{30} = \frac{5a - 3}{30}$

д) $\frac{2x + 3}{6} + \frac{x - 1}{8} = \frac{4(2x + 3) + 3(x - 1)}{24} = \frac{8x + 12 + 3x - 3}{24} = \frac{11x + 9}{24}$

е) $\frac{a - 3}{10} - \frac{2 - a}{15} = \frac{3(a - 3) - 2(2 - a)}{30} = \frac{3a - 9 - 4 + 2a}{30} = \frac{5a - 13}{30}$

Задание 517

Преобразуйте в алгебраическую дробь:
а) $\frac{1}{4x} - \frac{1}{3x}$;
б) $\frac{1}{m} + \frac{5}{4m}$;
в) $\frac{2}{p} + \frac{3}{pq}$;
г) $\frac{a}{xy} - \frac{b}{x}$;
д) $\frac{m}{n^2} - \frac{1}{mn}$;
е) $\frac{a}{3b^2} + \frac{8}{2ab}$.

Решение

а) $\frac{1}{4x} - \frac{1}{3x} = \frac{3 - 4}{12x} = \frac{-1}{12x} = -\frac{1}{12x}$

б) $\frac{1}{m} + \frac{5}{4m} = \frac{4 + 5}{4m} = \frac{9}{4m}$

в) $\frac{2}{p} + \frac{3}{pq} = \frac{2q + 3}{pq}$

г) $\frac{a}{xy} - \frac{b}{x} = \frac{a - by}{xy}$

д) $\frac{m}{n^2} - \frac{1}{mn} = \frac{m^2 - n}{mn^2}$

е) $\frac{a}{3b^2} + \frac{8}{2ab} = \frac{2a^2 + 24b}{6ab^2} = \frac{2(a^2 + 12b)}{6ab^2} = \frac{a^2 + 12b}{3ab^2}$

Задание 518

Преобразуйте в алгебраическую дробь:
а) $\frac{m}{ab} + \frac{m}{ac}$;
б) $\frac{2a}{mn} - \frac{5a}{mb}$;
в) $\frac{2a - 3b}{m} + \frac{4a - 5b^2}{mb}$;
г) $\frac{x - y}{xy} - \frac{x - z}{xz}$.

Решение

а) $\frac{m}{ab} + \frac{m}{ac} = \frac{cm + bm}{abc}$

б) $\frac{2a}{mn} - \frac{5a}{mb} = \frac{2ab - 5an}{bmn}$

в) $\frac{2a - 3b}{m} + \frac{4a - 5b^2}{mb} = \frac{b(2a - 3b) + 4a - 5b^2}{mb} = \frac{2ab - 3b^2 + 4a - 5b^2}{mb} = \frac{2ab + 4a - 8b^2}{mb}$

г) $\frac{x - y}{xy} - \frac{x - z}{xz} = \frac{z(x - y) - y(x - z)}{xyz} = \frac{xz - yz - xy + yz}{xyz} = \frac{xz - xy}{xyz} = \frac{x(z - y)}{xyz} = \frac{z - y}{yz}$

Задание 519

Преобразуйте в алгебраическую дробь:
а) $\frac{2}{x^2} - \frac{3}{x^3}$;
б) $\frac{7}{m^4} - \frac{3a}{m^2}$;
в) $\frac{1}{a^5b^3} + \frac{1}{ab^7}$;
г) $\frac{4}{x^4b^3} - \frac{3}{x^2b^5}$;
д) $\frac{3a}{x^7y^5z} - \frac{3b}{xy^4z^5}$;
е) $\frac{m^7n}{a^4b^3c^9} + \frac{3mn^2}{a^3b^6c^4}$.

Решение

а) $\frac{2}{x^2} - \frac{3}{x^3} = \frac{2x - 3}{x^3}$

б) $\frac{7}{m^4} - \frac{3a}{m^2} = \frac{7 - 3am^2}{m^4}$

в) $\frac{1}{a^5b^3} + \frac{1}{ab^7} = \frac{b^4 + a^4}{a^5b^7}$

г) $\frac{4}{x^4b^3} - \frac{3}{x^2b^5} = \frac{4b^2 - 3x^2}{x^4b^5}$

д) $\frac{3a}{x^7y^5z} - \frac{3b}{xy^4z^5} = \frac{3az^4 - 3bx^6y}{x^7y^5z^5}$

е) $\frac{m^7n}{a^4b^3c^9} + \frac{3mn^2}{a^3b^6c^4} = \frac{m^7nb^3 + 3mn^2ac^5}{a^4b^6c^9} = \frac{mn(b^3m^6 + 3ac^5n)}{a^4b^6c^9}$

Задание 520

Преобразуйте в алгебраическую дробь:
а) $\frac{1}{2a - 2} + \frac{2}{4a - 4}$;
б) $\frac{7a}{3x + 3} - \frac{a}{6x + 6}$;
в) $\frac{2m}{4m + 4n} + \frac{4n}{8m + 8n}$;
г) $\frac{2p}{10p - 10q} - \frac{3q}{15p - 15q}$;
д) $\frac{2x}{ax + bx} + \frac{3y}{ay + by}$;
е) $\frac{y}{ax - bx} - \frac{x}{ay - by}$;
ж) $\frac{1}{2x^2y - xy} + \frac{2}{y - 2xy}$;
з) $\frac{3}{3m^2n - 6mn^2} - \frac{2}{4mn - 2m^2}$;
и) $\frac{15}{10p^3q - 15p^2q^2} - \frac{6q}{9pq^3 - 6p^2q^2}$;
к) $\frac{3b}{2a^3b - 8a^2b^2} - \frac{5a}{12a^3b - 3a^4}$.

Решение

а) $\frac{1}{2a - 2} + \frac{2}{4a - 4} = \frac{1}{2(a - 1)} + \frac{2}{4(a - 1)} = \frac{2 + 2}{4(a - 1)} = \frac{4}{4(a - 1)} = \frac{1}{a - 1}$

б) $\frac{7a}{3x + 3} - \frac{a}{6x + 6} = \frac{7a}{3(x + 1)} - \frac{a}{6(x + 1)} = \frac{14a - a}{6(x + 1)} = \frac{13a}{6(x + 1)}$

в) $\frac{2m}{4m + 4n} + \frac{4n}{8m + 8n} = \frac{2m}{4(m + n)} + \frac{4n}{8(m + n)} = \frac{4m + 4n}{8(m + n)} = \frac{4(m + n)}{8(m + n)} = \frac{1}{2}$

г) $\frac{2p}{10p - 10q} - \frac{3q}{15p - 15q} = \frac{2p}{10(p - q)} - \frac{3q}{15(p - q)} = \frac{6p - 6q}{30(p - q)} = \frac{6(p - q)}{30(p - q)} = \frac{1}{5}$

д) $\frac{2x}{ax + bx} + \frac{3y}{ay + by} = \frac{2x}{x(a + b)} + \frac{3y}{y(a + b)} = \frac{2xy + 3xy}{xy(a + b)} = \frac{5xy}{xy(a + b)} = \frac{5}{a + b}$

е) $\frac{y}{ax - bx} - \frac{x}{ay - by} = \frac{y}{x(a - b)} - \frac{x}{y(a - b)} = \frac{y^2 - x^2}{xy(a - b)}$

ж) $\frac{1}{2x^2y - xy} + \frac{2}{y - 2xy} = \frac{1}{xy(2x - 1} + \frac{2}{y(1 - 2x)} = \frac{1}{xy(2x - 1} - \frac{2}{y(2x - 1)} = \frac{1 - 2x}{xy(2x - 1} = -\frac{2x - 1}{xy(2x - 1} = -\frac{1}{xy}$

з) $\frac{3}{3m^2n - 6mn^2} - \frac{2}{4mn - 2m^2} = \frac{3}{3mn(m - 2n)} - \frac{2}{2m(2n - m)} = \frac{3}{3mn(m - 2n)} + \frac{2}{2m(m - 2n)} = \frac{6 + 6n}{6mn(m - 2n)} = \frac{6(1 + n)}{6mn(m - 2n)} = \frac{1 + n}{mn(m - 2n)}$

и) $\frac{15}{10p^3q - 15p^2q^2} - \frac{6q}{9pq^3 - 6p^2q^2} = \frac{15}{5p^2q(2p - 3q)} - \frac{6q}{3pq^2(3q - 2p)} = \frac{15}{5p^2q(2p - 3q)} + \frac{6q}{3pq^2(2p - 3q)} = \frac{45q + 30pq}{15p^2q^2(2p - 3q)} = \frac{15q(3 + 2p)}{15p^2q^2(2p - 3q)} = \frac{3 + 2p}{p^2q(2p - 3q)}$

к) $\frac{3b}{2a^3b - 8a^2b^2} - \frac{5a}{12a^3b - 3a^4} = \frac{3b}{2a^2b(a - 4b)} - \frac{5a}{3a^3(4b - a)} = \frac{3b}{2a^2b(a - 4b)} + \frac{5a}{3a^3(a - 4b)} = \frac{9ab + 10ab}{6a^3b(a - 4b)} = \frac{19ab}{6a^3b(a - 4b)} = \frac{19}{6a^2(a - 4b)}$

Задание 521

Преобразуйте в алгебраическую дробь:
а) $\frac{2a}{a^2 - 9} + \frac{3}{a - 3}$;
б) $\frac{5}{m + n} - \frac{4n}{m^2 - n^2}$;
в) $\frac{x}{4 - 9x^2} + \frac{1}{3x + 2}$;
г) $\frac{1}{2p + 4q} - \frac{q}{4q^2 - p^2}$;
д) $\frac{1}{a^2 + ab + b^2} + \frac{b}{a^3 - b^3}$;
е) $\frac{m^2 + n^2}{m^3 + n^3} - \frac{1}{2(m + n)}$;
ж) $\frac{x^2 - 2xy}{(x - 2y)^3} + \frac{1}{2y - x}$;
з) $\frac{2(p + q)}{p^3 - q^3} + \frac{3}{q^2 - p^2}$.

Решение

а) $\frac{2a}{a^2 - 9} + \frac{3}{a - 3} = \frac{2a}{(a - 3)(a + 3)} + \frac{3}{a - 3} = \frac{2a + 3(a + 3)}{(a - 3)(a + 3)} = \frac{2a + 3a + 9}{(a - 3)(a + 3)} = \frac{5a + 9}{a^2 - 9}$

б) $\frac{5}{m + n} - \frac{4n}{m^2 - n^2} = \frac{5}{m + n} - \frac{4n}{(m - n)(m + n)} = \frac{5(m - n) - 4n}{(m - n)(m + n)} = \frac{5m - 5n - 4n}{(m - n)(m + n)} = \frac{5m - 9n}{m^2 - n^2}$

в) $\frac{x}{4 - 9x^2} + \frac{1}{3x + 2} = \frac{x}{(2 - 3x)(2 + 3x)} + \frac{1}{3x + 2} = \frac{x + 2 - 3x}{(2 - 3x)(2 + 3x)} = \frac{2 - 2x}{(2 - 3x)(2 + 3x)} = \frac{2(1 - x)}{4 - 9x^2}$

г) $\frac{1}{2p + 4q} - \frac{q}{4q^2 - p^2} = \frac{1}{2(p + 2q)} - \frac{q}{(2q - p)(2q + p)} = \frac{2q - p - 2q}{2(2q - p)(2q + p)} = \frac{-p}{2(2q - p)(2q + p)} = -\frac{p}{2(4q^2 - p^2)}$

д) $\frac{1}{a^2 + ab + b^2} + \frac{b}{a^3 - b^3} = \frac{1}{a^2 + ab + b^2} + \frac{b}{(a - b)(a^2 + ab + b^2)} = \frac{a - b + b}{(a - b)(a^2 + ab + b^2)} = \frac{a}{a^3 - b^3}$

е) $\frac{m^2 + n^2}{m^3 + n^3} - \frac{1}{2(m + n)} = \frac{m^2 + n^2}{(m + n)(m^2 - mn + n^2)} - \frac{1}{2(m + n)} = \frac{2(m^2 + n^2) - (m^2 - mn + n^2)}{2(m + n)(m^2 - mn + n^2)} = \frac{2m^2 + 2n^2 - m^2 + mn - n^2}{2(m + n)(m^2 - mn + n^2)} = \frac{m^2 + n^2 + mn}{2(m^3 + n^3)}$

ж) $\frac{x^2 - 2xy}{(x - 2y)^3} + \frac{1}{2y - x} = \frac{x(x - 2y)}{(x - 2y)^3} + \frac{1}{2y - x} = \frac{x}{(x - 2y)^2} + \frac{1}{2y - x} = \frac{x}{(2y - x)^2} + \frac{1}{2y - x} = \frac{x + 2y - x}{(2y - x)^2} = \frac{2y}{(x - 2y)^2}$

з) $\frac{2(p + q)}{p^3 - q^3} + \frac{3}{q^2 - p^2} = \frac{2(p + q)}{(p - q)(p^2 + pq + q^2)} + \frac{3}{(q - p)(q + p)} = \frac{2(p + q)}{(p - q)(p^2 + pq + q^2)} - \frac{3}{(p - q)(p + q)} = \frac{2(p + q)(p + q) - 3(p^2 + pq + q^2)}{(p - q)(p + q)(p^2 + pq + q^2)} = \frac{2(p + q)^2 - 3p^2 - 3pq - 3q^2}{(p - q)(p + q)(p^2 + pq + q^2)} = \frac{2(p^2 + 2pq + q) - 3p^2 - 3pq - 3q^2}{(p - q)(p + q)(p^2 + pq + q^2)} = \frac{2p^2 + 4pq + 2q^2 - 3p^2 - 3pq - 3q^2}{(p - q)(p + q)(p^2 + pq + q^2)} = \frac{-p^2 + pq - q^2}{(p - q)(p + q)(p^2 + pq + q^2)} = \frac{-p^2 + pq - q^2}{(p^3 - q^3)(p + q)}$