Решебник по алгебре 8 класс Мерзляк ФГОС Задание 156

 

\[\boxed{\text{156\ (156).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\frac{3a^{4}b^{3}}{10c^{5}} \cdot \frac{4b^{4}c^{2}}{27a^{7}}\ :\frac{5b^{7}}{9a^{3}c^{3}} =\]

\[= \frac{3a^{4}b^{3} \cdot 4b^{4}c^{2} \cdot 9a^{3}c^{3}}{10c^{5} \cdot 27a^{7} \cdot 5b^{7}} = \frac{2}{25}\]

\[2)\frac{3a^{2}}{2b^{2}c^{2}}\ :\frac{7c^{8}}{6b^{3}}\ :\frac{9ab}{14c^{12}} =\]

\[= \frac{3a^{2} \cdot 6b^{3} \cdot 14c^{12}}{2b^{2}c^{2} \cdot 7c^{8} \cdot 9ab} = \frac{2ac^{2}}{1} =\]

\[= 2ac^{2}\]

\[3)\ \left( \frac{5a^{3}}{b^{4}} \right)^{4} \cdot \frac{b^{18}}{50a^{16}} =\]

\[= \frac{25 \cdot 25 \cdot a^{12} \cdot b^{18}}{b^{16} \cdot 50a^{16}} = \frac{25b^{2}}{2a^{4}}\]

\[4)\ \left( \frac{3x^{7}}{y^{10}} \right)^{4}:\left( \frac{3x^{6}}{y^{8}} \right)^{3} =\]

\[= \frac{3^{4} \cdot x^{28} \cdot y^{24}}{y^{40} \cdot 3^{3} \cdot x^{18}} = \frac{3x^{10}}{y^{16}}\]

\[\boxed{\text{156.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{4 - a}{8a^{3}} \cdot \frac{12a^{5}}{a^{2} - 16} =\]

\[= \frac{(4 - a) \cdot 12a^{5}}{8a^{3} \cdot (a - 4)(a + 4)} =\]

\[= \frac{- 3a^{2}}{2 \cdot (a + 4)}\]

\[2)\ \frac{4c - d}{c^{2} + cd} \cdot \frac{2c^{2} - 2d^{2}}{4c^{2} - cd} =\]

\[= \frac{(4c - d)(c - d)(c + d) \cdot 2}{c(c + d) \cdot c \cdot (4c - d)} =\]

\[= \frac{2 \cdot (c - d)}{c^{2}}\]

\[3)\ \frac{b^{2} - 6b + 9}{b^{2} - 3b + 9} \cdot \frac{b^{3} + 27}{5b - 15} =\]

\[= \frac{(b - 3)^{2} \cdot (b + 3) \cdot \left( b^{2} - 3b + 9 \right)}{\left( b^{2} - 3b + 9 \right) \cdot 5 \cdot (b - 3)} =\]

\[= \frac{b^{2} - 9}{5}\]

\[4)\ \frac{a^{3} - 16a}{3a^{2}b} \cdot \frac{12ab^{2}}{4a + 16} =\]

\[= \frac{a(a - 4)(a + 4) \cdot 12ab^{2}}{3a^{2}b \cdot 4 \cdot (a + 4)} =\]

\[= b(a - 4)\]

\[5)\ \frac{a^{3} + b^{3}}{a^{2} - b^{2}} \cdot \frac{7a - 7b}{a^{2} - ab + b^{2}} =\]

\[= \frac{(a + b)\left( a^{2} - ab + b^{2} \right) \cdot 7(a - b)}{(a - b)(a + b)\left( a^{2} - ab + b^{2} \right)} = 7\]

\[6)\ \frac{x^{2} - 9}{x + y} \cdot \frac{5x + 5y}{x^{2} - 3x} =\]

\[= \frac{(x - 3)(x + 3) \cdot 5 \cdot (x + y)}{(x + y) \cdot x \cdot (x - 3)} =\]

\[= \frac{5x + 15}{x}\text{\ \ }\]

\[7)\ \frac{m + 2n}{2 - 3m}\ :\frac{m^{2} + 4mn + 4n^{2}}{3m^{2} - 2m} =\]

\[= \frac{(m + 2n) \cdot m \cdot (3m - 2)}{(2 - 3m)(m + 2n)^{2}} =\]

\[= \frac{- m}{m + 2n}\]

\[8)\ \frac{a^{3} + 8}{16 - a^{4}}\ :\frac{a^{2} - 2a + 4}{a^{2} + 4} =\]

\[= \frac{(a + 2)\left( a^{2} - 2a + 4 \right)\left( a^{2} + 4 \right)}{\left( 4 - a^{2} \right)\left( 4 + a^{2} \right)\left( a^{2} - 2a + 4 \right)} =\]

\[= \frac{a + 2}{4 - a^{2}} =\]

\[= \frac{(a + 2)}{(2 - a)(2 + a)} = \frac{1}{2 - a}\]

\[9)\ \frac{x^{2} - 12x + 36}{3x + 21} \cdot \frac{x^{2} - 49}{4x - 24} =\]

\[= \frac{(x - 6)^{2} \cdot (x - 7)(x + 7)}{3(x + 7) \cdot 4 \cdot (x - 6)} =\]

\[= \frac{(x - 6)(x - 7)}{12}\]

\[10)\ \frac{3a + 15b}{a^{2} - 81b^{2}}\ :\]

\[:\frac{4a + 20b}{a^{2} - 18ab + 81b^{2}} =\]

\[= \frac{3(a + 5b)(a - 9b)^{2}}{(a - 9b)(a + 9b) \cdot 4 \cdot (a + 5b)} =\]

\[= \frac{3 \cdot (a - 9b)}{4 \cdot (a + 9b)}\]