01Математика - 9 класс. Алгебра - Нахождение значений буквенных выражений

Задание

Найдите значение выражения

при \(\displaystyle x=5\small,\) \(\displaystyle y=128\small.\)

Таблица степеней

Информация

Таблица степеней

		С т е п е н и
		\(\displaystyle \bf \color{blue}{2}\)	\(\displaystyle \bf \color{blue}{3}\)	\(\displaystyle \bf \color{blue}{4}\)	\(\displaystyle \bf \color{blue}{5}\)	\(\displaystyle \bf \color{blue}{6}\)	\(\displaystyle \bf \color{blue}{7}\)	\(\displaystyle \bf \color{blue}{8}\)
Ч и с л а	\(\displaystyle \bf \color{blue}{2}\)	\(\displaystyle 4\)	\(\displaystyle 8\)	\(\displaystyle 16\)	\(\displaystyle 32\)	\(\displaystyle 64\)	\(\displaystyle 128\)	\(\displaystyle 256\)
	\(\displaystyle \bf \color{blue}{3}\)	\(\displaystyle 9\)	\(\displaystyle 27\)	\(\displaystyle 81\)	\(\displaystyle 243\)	\(\displaystyle 729\)	\(\displaystyle 2187\)	\(\displaystyle 6561\)
	\(\displaystyle \bf \color{blue}{4}\)	\(\displaystyle 16\)	\(\displaystyle 64\)	\(\displaystyle 256\)	\(\displaystyle 1024\)	\(\displaystyle 4096\)	\(\displaystyle 16384\)	\(\displaystyle \ldots\)
	\(\displaystyle \bf \color{blue}{5}\)	\(\displaystyle 25\)	\(\displaystyle 125\)	\(\displaystyle 625\)	\(\displaystyle 3125\)	\(\displaystyle 15625\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)
	\(\displaystyle \bf \color{blue}{6}\)	\(\displaystyle 36\)	\(\displaystyle 216\)	\(\displaystyle 1296\)	\(\displaystyle 7776\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)
	\(\displaystyle \bf \color{blue}{7}\)	\(\displaystyle 49\)	\(\displaystyle 343\)	\(\displaystyle 2401\)	\(\displaystyle 16807\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)
	\(\displaystyle \bf \color{blue}{8}\)	\(\displaystyle 64\)	\(\displaystyle 512\)	\(\displaystyle 4096\)	\(\displaystyle 32768\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)
	\(\displaystyle \bf \color{blue}{9}\)	\(\displaystyle 81\)	\(\displaystyle 729\)	\(\displaystyle 6561\)	\(\displaystyle 59049\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)
	\(\displaystyle \bf \color{blue}{10}\)	\(\displaystyle 100\)	\(\displaystyle 1000\)	\(\displaystyle 10000\)	\(\displaystyle 100000\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)	\(\displaystyle \ldots\)

Решение

Сначала упростим выражение.

Получаем:

\(\displaystyle \frac{x^{\frac{2}{3}}+5x^{\frac{1}{3}}y^{\frac{1}{7}}-2x^{\frac{1}{3}}-10y^{\frac{1}{7}}}{x^{\frac{2}{3}}-2x^{\frac{1}{3}}}\cdot \frac{3x^{\frac{1}{3}}y^{\frac{1}{7}}}{x^{\frac{1}{3}}+5y^{\frac{1}{7}}}=\frac{(x^{\frac{2}{3}}+5x^{\frac{1}{3}}y^{\frac{1}{7}}-2x^{\frac{1}{3}}-10y^{\frac{1}{7}})\cdot 3x^{\frac{1}{3}}y^{\frac{1}{7}}}{(x^{\frac{2}{3}}-2x^{\frac{1}{3}})\cdot(x^{\frac{1}{3}}+5y^{\frac{1}{7}})}{\small .}\)

Разложим выражения \(\displaystyle x^{\frac{2}{3}}+5x^{\frac{1}{3}}y^{\frac{1}{7}}-2x^{\frac{1}{3}}-10y^{\frac{1}{7}}\) и \(\displaystyle x^{\frac{2}{3}}-2x^{\frac{1}{3}}\) на множители:

\(\displaystyle \,\)•\(\displaystyle \color{black}{x^{\frac{2}{3}}+5x^{\frac{1}{3}}y^{\frac{1}{7}}-2x^{\frac{1}{3}}-10y^{\frac{1}{7}}=(x^{\frac{1}{3}}+5y^{\frac{1}{7}})(x^{\frac{1}{3}}-2)\small,} \)

Воспользуемся методом группировки:

\(\displaystyle \begin{aligned}& x^{\frac{2}{3}}+5x^{\frac{1}{3}}y^{\frac{1}{7}}-2x^{\frac{1}{3}}-10y^{\frac{1}{7}}=(x^{\frac{2}{3}}+5x^{\frac{1}{3}}y^{\frac{1}{7}})-(2x^{\frac{1}{3}}+10y^{\frac{1}{7}}) =\\ \\& \qquad\qquad\qquad = x^{\frac{1}{3}}(x^{\frac{1}{3}}+5y^{\frac{1}{7}})-2(x^{\frac{1}{3}}+5y^{\frac{1}{7}})=(x^{\frac{1}{3}}+5y^{\frac{1}{7}})(x^{\frac{1}{3}}-2){\small .}\end{aligned} \)

\(\displaystyle x^{\frac{2}{3}}-2x^{\frac{1}{3}}=x^{\frac{1}{3}}(x^{\frac{1}{3}}-2){\small .}\)

Подставляя, получаем:

\(\displaystyle \frac{(x^{\frac{2}{3}}+5x^{\frac{1}{3}}y^{\frac{1}{7}}-2x^{\frac{1}{3}}-10y^{\frac{1}{7}})\cdot 3x^{\frac{1}{3}}y^{\frac{1}{7}}}{(x^{\frac{2}{3}}-2x^{\frac{1}{3}})\cdot (x^{\frac{1}{3}}+5y^{\frac{1}{7}})} = \frac{3x^{\frac{1}{3}}y^{\frac{1}{7}}(x^{\frac{1}{3}}+5y^{\frac{1}{7}})(x^{\frac{1}{3}}-2)}{x^{\frac{1}{3}}(x^{\frac{1}{3}}-2)(x^{\frac{1}{3}}+5y^{\frac{1}{7}})} {\small .}\)

Сокращая, получаем:

\(\displaystyle \begin{aligned}\frac{3y^{\frac{1}{7}}\,\cancel{\color{orange}{\,\,x^{\frac{1}{3}}}}{\cancel{\color{blue}{(x^{\frac{1}{3}}+5y^{\frac{1}{7}})}}}\cancel{\color{green}{(x^{\frac{1}{3}}-2)}}}{{\cancel{\color{orange}{\,\,x^{\frac{1}{3}}}} \cancel{\color{green}{(x^{\frac{1}{3}}-2)}} \cancel{\color{blue}{(x^{\frac{1}{3}}+5y^{\frac{1}{7}})}}}}=3y^{\frac{1}{7}}{\small .}\end{aligned}\)

При \(\displaystyle x=5\small,\) \(\displaystyle y=128\small\) получим

\(\displaystyle 3\cdot 128^{\frac{1}{7}}=3\cdot \sqrt[7]{128^1}=3\cdot \sqrt[7]{128}=3\cdot 2=6{\small .}\)

Ответ: \(\displaystyle 6\small.\)

Теория: 19 Нахождение значений буквенных выражений

Таблица степеней

С т е п е н и

Ч и с л а

Ч
и
с
л
а