Решебник по алгебре 8 класс Макарычев ФГОС Задание 84

 

\[\boxed{\text{84\ (84).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

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

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

\[= \frac{6}{x \cdot (x - 2)}\]

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

\[= \frac{4a^{2} + 1}{4a^{2} - 1}\]

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

\[= \frac{2p}{9p^{2} - 1}\]

\[\boxed{\text{84.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\mathbf{Знаменатель\ целых\ чисел\ }\]

\[\mathbf{равен\ 1.}\]

\[\textbf{а)}\ 1^{\backslash 20} - \frac{a^{\backslash 4}}{5} - \frac{b^{\backslash 5}}{4} =\]

\[= \frac{20 - 4a - 5b}{20}\]

\[\textbf{б)}\ 12^{\backslash ab} - \frac{1^{\backslash b}}{a} - \frac{1^{\backslash a}}{b} =\]

\[= \frac{12ab - b - a}{\text{ab}}\]

\[\textbf{в)}\ \frac{a - 2^{\backslash 3}}{2} - 1^{\backslash 6} - \frac{a - 3^{\backslash 2}}{3} =\]

\[= \frac{3a - 6 - 6 - 2a + 6}{6} = \frac{a - 6}{6}\]

\[\textbf{г)}\ 4a^{\backslash 12} - \frac{a - 1^{\backslash 3}}{4} - \frac{a + 2^{\backslash 4}}{3} =\]

\[= \frac{48a - 3a + 3 - 4a - 8}{12} =\]

\[= \frac{41a - 5}{12}\ \]

\[\textbf{д)}\ \frac{a + b}{4} - a + b =\]

\[= \frac{a + b}{4} - (a - b)^{\backslash 4} =\]

\[= \frac{a + b - 4a + 4b}{4} = \frac{5b - 3a}{4}\]

\[\textbf{е)}\ a + b - \frac{a^{2} + b^{2}}{a} =\]

\[= (a + b)^{\backslash a} - \frac{a^{2} + b^{2}}{a} =\]

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