Задание 522

Преобразуйте в алгебраическую дробь:
а) $3 - \frac{7}{m - 2}$;
б) $1 - \frac{x - y}{x + y}$;
в) $\frac{(a + b)^2}{b} - 2a$;
г) $\frac{(a - b)^2}{2a} + b$;
д) $a + b - \frac{a^2 + b^2}{a - b}$;
е) $\frac{a^2 + b^2}{a + b} + a - b$.

Решение

а) $3 - \frac{7}{m - 2} = \frac{3(m - 2) - 7}{m - 2} = \frac{3m - 6 - 7}{m - 2} = \frac{3m - 13}{m - 2}$

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

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

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

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

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

Задание 523

Преобразуйте в алгебраическую дробь:
а) $\frac{a}{b} * \frac{c}{d}$;
б) $\frac{x}{y} : \frac{a}{b}$;
в) $\frac{4a}{7b} * \frac{21}{a}$;
г) $\frac{5}{8} : \frac{15q}{16p}$;
д) $\frac{5ax}{6by} * \frac{3x}{5y}$;
е) $\frac{7}{a} * \frac{5ax}{14by}$;
ж) $\frac{8a^2y}{5bx} : \frac{3ay}{4b^2x}$;
з) $\frac{25x^2y^3}{36ab} : \frac{35x^3y}{24b^2}$.

Решение

а) $\frac{a}{b} * \frac{c}{d} = \frac{aс}{bd}$

б) $\frac{x}{y} : \frac{a}{b} = \frac{x}{y} * \frac{b}{a} = \frac{bx}{ay}$

в) $\frac{4a}{7b} * \frac{21}{a} = \frac{4}{b} * \frac{3}{1} = \frac{12}{b}$

г) $\frac{5}{8} : \frac{15q}{16p} = \frac{5}{8} * \frac{16p}{15q} = \frac{2p}{3q}$

д) $\frac{5ax}{6by} * \frac{3x}{5y} = \frac{ax}{2by} * \frac{x}{y} = \frac{ax^2}{2by^2}$

е) $\frac{7}{a} * \frac{5ax}{14by} = \frac{5x}{2by}$

ж) $\frac{8a^2y}{5bx} : \frac{3ay}{4b^2x} = \frac{8a^2y}{5bx} * \frac{4b^2x}{3ay} = \frac{8a}{5} * \frac{4b}{3} = \frac{32ab}{15}$

з) $\frac{25x^2y^3}{36ab} : \frac{35x^3y}{24b^2} = \frac{25x^2y^3}{36ab} * \frac{24b^2}{35x^3y} = \frac{5y^2}{3a} * \frac{2b}{7x} = \frac{10by^2}{21ax}$

Задание 524

Преобразуйте в алгебраическую дробь:
а) $a * \frac{a}{b}$;
б) $\frac{a}{x} : a$;
в) $\frac{a}{7x} * 5x$;
г) $ab : \frac{a}{b}$;
д) $8a : \frac{20a^2b}{3x}$;
е) $18p^3 * \frac{5x}{9p^2}$.

Решение

а) $a * \frac{a}{b} = \frac{a}{1} * \frac{a}{b} = frac{a^2}{b}$

б) $\frac{a}{x} : a = \frac{a}{x} * \frac{1}{a} = \frac{1}{x}$

в) $\frac{a}{7x} * 5x = \frac{a}{7x} * \frac{5x}{1} = \frac{5a}{7}$

г) $ab : \frac{a}{b} = \frac{ab}{1} * \frac{b}{a} = b * b = b^2$

д) $8a : \frac{20a^2b}{3x} = \frac{8a}{1} * \frac{3x}{20a^2b} = \frac{2}{1} * \frac{3x}{5ab} = \frac{6x}{5ab}$

е) $18p^3 * \frac{5x}{9p^2} = \frac{18p^3}{1} * \frac{5x}{9p^2} = \frac{2p}{1} * \frac{5x}{1} = 10px$

Задание 525

Преобразуйте в алгебраическую дробь:
а) $\frac{a + 1}{7x} * \frac{2x}{a + 1}$;
б) $\frac{2m}{m - n} : \frac{3mn}{m - n}$;
в) $\frac{4p}{p - 3} * \frac{p - 3}{2p^2}$;
г) $\frac{x + y}{8a} : \frac{x + y}{16a^2b}$;
д) $\frac{2x + 2y}{3} * \frac{6}{x + y}$;
е) $\frac{4a}{a^2b} : \frac{5ab}{3a - 3b}$;
ж) $\frac{m - 3n}{6m} * \frac{3mn}{4m - 12n}$;
з) $\frac{2p - 4q}{3p^2} : \frac{3p - 6q}{4pq}$;
и) $\frac{ax - ay}{cd} * \frac{cx + cy}{x - y}$;
к) $\frac{mk}{am - an} : \frac{ka - k}{2m - 2n}$.

Решение

а) $\frac{a + 1}{7x} * \frac{2x}{a + 1} = \frac{1}{7} * \frac{2}{1} = \frac{2}{7}$

б) $\frac{2m}{m - n} : \frac{3mn}{m - n} = \frac{2m}{m - n} * \frac{m - n}{3mn} = \frac{2}{1} * \frac{1}{3n} = \frac{2}{3n}$

в) $\frac{4p}{p - 3} * \frac{p - 3}{2p^2} = \frac{2}{1} * \frac{1}{p} = \frac{2}{p}$

г) $\frac{x + y}{8a} : \frac{x + y}{16a^2b} = \frac{x + y}{8a} * \frac{16a^2b}{x + y} = \frac{1}{1} * \frac{2ab}{1} = 2ab$

д) $\frac{2x + 2y}{3} * \frac{6}{x + y} = \frac{2(x + y)}{3} * \frac{6}{x + y} = \frac{2}{1} * \frac{2}{1} = 4$

е) $\frac{4a}{a^2b} : \frac{5ab}{3a - 3b} = \frac{4a}{a^2b} * \frac{3a - 3b}{5ab} = \frac{4a}{a^2b} * \frac{3(a - b)}{5ab} = \frac{4}{a^2b} * \frac{3(a - b)}{5b} = \frac{12(a - b)}{5a^2b^2}$

ж) $\frac{m - 3n}{6m} * \frac{3mn}{4m - 12n} = \frac{m - 3n}{6m} * \frac{3mn}{4(m - 3n)} = \frac{1}{2} * \frac{n}{4} = \frac{n}{8}$

з) $\frac{2p - 4q}{3p^2} : \frac{3p - 6q}{4pq} = \frac{2p - 4q}{3p^2} * \frac{4pq}{3p - 6q} = \frac{2(p - 2q)}{3p^2} * \frac{4pq}{3(p - 2q)} = \frac{2}{3p} * \frac{4q}{3} = \frac{8q}{9p}$

и) $\frac{ax - ay}{cd} * \frac{cx + cy}{x - y} = \frac{a(x - y)}{cd} * \frac{c(x + y)}{x - y} = \frac{a}{d} * \frac{x + y}{1} = \frac{a(x + y)}{d}$

к) $\frac{mk}{am - an} : \frac{ka - k}{2m - 2n} = \frac{mk}{am - an} * \frac{2m - 2n}{ka - k} = \frac{mk}{a(m - n)} * \frac{2(m - n)}{k(a - 1)} = \frac{m}{a} * \frac{2}{a - 1} = \frac{2m}{a(a - 1)}$

Задание 526

Преобразуйте в алгебраическую дробь:
а) $\frac{a^2 - b^2}{2a^2b} * \frac{4ab^2}{a + b}$;
б) $\frac{(x - y)^2}{3x^2y^3} : \frac{x - y}{6xy^2}$;
в) $\frac{mn - m^2}{2m} * \frac{8n}{n^2 - m^2}$;
г) $\frac{2a - 4}{b + 1} : \frac{a^2 - 4}{(b + 1)^2}$;
д) $\frac{x + y}{x - y} * \frac{x^2 - xy}{2x^2 - 2y^2}$;
е) $\frac{16 - m^2}{m^2 - 3m} : \frac{m^2 + 4m}{m^2 - 9}$.

Решение

а) $\frac{a^2 - b^2}{2a^2b} * \frac{4ab^2}{a + b} = \frac{(a - b)(a + b)}{2a^2b} * \frac{4ab^2}{a + b} = \frac{a - b}{a} * \frac{2b}{1} = \frac{2b(a - b)}{a}$

б) $\frac{(x - y)^2}{3x^2y^3} : \frac{x - y}{6xy^2} = \frac{(x - y)^2}{3x^2y^3} * \frac{6xy^2}{x - y} = \frac{x - y}{xy} * \frac{2}{1} = \frac{2(x - y)}{xy}$

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

г) $\frac{2a - 4}{b + 1} : \frac{a^2 - 4}{(b + 1)^2} = \frac{2a - 4}{b + 1} * \frac{(b + 1)^2}{a^2 - 4} = \frac{2(a - 2)}{b + 1} * \frac{(b + 1)^2}{(a - 2)(a + 2)} = \frac{2}{1} * \frac{b + 1}{a + 2} = \frac{2(b + 1)}{a + 2}$

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

е) $\frac{16 - m^2}{m^2 - 3m} : \frac{m^2 + 4m}{m^2 - 9} = \frac{16 - m^2}{m^2 - 3m} * \frac{m^2 - 9}{m^2 + 4m} = \frac{(4 - m)(4 + m)}{m(m - 3)} * \frac{(m - 3)(m + 3)}{m(m + 4)} = \frac{4 - m}{m} * \frac{m + 3}{m} = \frac{(4 - m)(m + 3)}{m^2}$

Задание 527

Преобразуйте в алгебраическую дробь:
а) $\frac{p^2 - q^2}{p^2} * \frac{pq + q^2}{(p + q)^2}$;
б) $\frac{a^2 - 9b^2}{a^2 - ab} : \frac{a^2 + 3ab}{a - b}$;
в) $\frac{3x^2 - 3y^2}{x^2 + xy} * \frac{x + y}{6x - 6y}$;
г) $\frac{m^2 - n^2}{(m + n)^2} : \frac{4m - 4n}{3m + 3n}$.

Решение

а) $\frac{p^2 - q^2}{p^2} * \frac{pq + q^2}{(p + q)^2} = \frac{(p - q)(p + q)}{p^2} * \frac{q(p + q)}{(p + q)^2} = \frac{(p - q)(p + q)}{p^2} * \frac{q}{p + q} = \frac{p - q}{p^2} * \frac{q}{1} = \frac{q(p - q)}{p^2}$

б) $\frac{a^2 - 9b^2}{a^2 - ab} : \frac{a^2 + 3ab}{a - b} = \frac{a^2 - 9b^2}{a^2 - ab} * \frac{a - b}{a^2 + 3ab} = \frac{(a - 3b)(a + 3b)}{a(a - b)} * \frac{a - b}{a(a + 3b)} = \frac{a - 3b}{a} * \frac{1}{a} = \frac{a - 3b}{a^2}$

в) $\frac{3x^2 - 3y^2}{x^2 + xy} * \frac{x + y}{6x - 6y} = \frac{3(x^2 - y^2)}{x(x + y)} * \frac{x + y}{6(x - y)} = \frac{3(x - y)(x + y)}{x(x + y)} * \frac{x + y}{6(x - y)} = \frac{3(x - y)}{x} * \frac{x + y}{6(x - y)} = \frac{1}{x} * \frac{x + y}{2} = \frac{x + y}{2x}$

г) $\frac{m^2 - n^2}{(m + n)^2} : \frac{4m - 4n}{3m + 3n} = \frac{m^2 - n^2}{(m + n)^2} * \frac{3m + 3n}{4m - 4n} = \frac{(m - n)(m + n)}{(m + n)^2} * \frac{3(m + n)}{4(m - n)} = \frac{m - n}{m + n} * \frac{3(m + n)}{4(m - n)} = \frac{3}{4}$

Задание 528

Преобразуйте в алгебраическую дробь:
а) $\frac{m^3 + n^3}{2m} * \frac{4mn}{m^2 - mn + n^2}$;
б) $\frac{2a}{a^3 - b^3} : \frac{6ab}{a^2 - b^2}$;
в) $\frac{m^3 - n^3}{m^3 + n^3} : \frac{(m - n)^2}{m^2 - n^2}$;
г) $\frac{x^2 + xy}{6x^2 - 6y^2} * \frac{3x^3 + 3y^3}{x^2 - xy}$;
д) $\frac{p^2 - 4q^2}{(p + 2q)^2} : \frac{p^3 - 8q^3}{4q^2 + 2pq + p^2}$;
е) $\frac{12a^2 + 6ab}{8a^3 - b^3} * \frac{4a^2 + 2ab + b^2}{3a^2 - 6ab}$.

Решение

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

б) $\frac{2a}{a^3 - b^3} : \frac{6ab}{a^2 - b^2} = \frac{2a}{a^3 - b^3} * \frac{a^2 - b^2}{6ab} = \frac{2a}{(a - b)(a^2 + ab + b^2)} * \frac{(a - b)(a + b)}{6ab} = \frac{1}{a^2 + ab + b^2} * \frac{a + b}{3b} = \frac{a + b}{3b(a^2 + ab + b^2)}$

в) $\frac{m^3 - n^3}{m^3 + n^3} : \frac{(m - n)^2}{m^2 - n^2} = \frac{m^3 - n^3}{m^3 + n^3} * \frac{m^2 - n^2}{(m - n)^2} = \frac{(m - n)(m^2 + mn + n^2)}{(m + n)(m^2 - mn + n^2)} * \frac{(m - n)(m + n)}{(m - n)^2} = \frac{(m - n)(m^2 + mn + n^2)}{(m + n)(m^2 - mn + n^2)} * \frac{m + n}{m - n} = \frac{m^2 + mn + n^2}{m^2 - mn + n^2}$

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

д) $\frac{p^2 - 4q^2}{(p + 2q)^2} : \frac{p^3 - 8q^3}{4q^2 + 2pq + p^2} = \frac{p^2 - 4q^2}{(p + 2q)^2} * \frac{4q^2 + 2pq + p^2}{p^3 - 8q^3} = \frac{(p - 2q)(p + 2q)}{p^2 + 4pq + 4q^2} * \frac{4q^2 + 2pq + p^2}{(p - 2q)(p^2 + 2pq + 4q^2)} = \frac{p + 2q}{1} * \frac{1}{p^2 + 2pq + 4q^2} = \frac{p + 2q}{(p + 2q)^2} = \frac{1}{p + 2q}$

е) $\frac{12a^2 + 6ab}{8a^3 - b^3} * \frac{4a^2 + 2ab + b^2}{3a^2 - 6ab} = \frac{6a(2a + b)}{(2a - b)(4a^2 + 2ab + b^2)} * \frac{4a^2 + 2ab + b^2}{3a(a - 2b)} = \frac{2(2a + b)}{2a - b} * \frac{1}{a - 2b} = \frac{2(2a + b)}{(2a - b)(2a - b)}$