Страница 245 номер 924, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Освободитесь от иррациональности в знаменателе дроби: 1) \(\frac{a}{2a - \sqrt{7b}}\); 2) \(\frac{p - 3}{\sqrt{4 - p} - 1}\); 3) \(\frac{m}{m + \sqrt{n}}\); 4) \(\frac{1}{\sqrt{a - 3} + 2}\); 5) \(\frac{7}{3 - \sqrt{b + 2}}\).
1)
\(\dfrac{a}{2a - \sqrt{7b}} = \dfrac{a(2a + \sqrt{7b})}{\left( 2a - \sqrt{7b} \right)(2a + \sqrt{7b})} = \dfrac{2a^{2} + a\sqrt{7b}}{4a^{2} - 7b}\)
2)
\(\dfrac{p - 3}{\sqrt{4 - p} - 1} = \dfrac{(p - 3)\left( \sqrt{4 - p} + 1 \right)}{\left( \sqrt{4 - p} - 1 \right)\left( \sqrt{4 - p} + 1 \right)} =\)
\(= \dfrac{(p - 3)(\sqrt{4 - p} + 1)}{4 - p - 1} = \dfrac{(p - 3)(\sqrt{4 - p} + 1)}{3 - p} =\)
\(= - (\sqrt{4 - p} + 1) = - \sqrt{4 - p} - 1\)
3)
\(\dfrac{m}{m + \sqrt{n}} = \dfrac{m\left( m - \sqrt{n} \right)}{\left( m + \sqrt{n} \right)(m - \sqrt{n})} = \dfrac{m(m - \sqrt{n})}{m^{2} - n}\)
4)
\(\dfrac{1}{\sqrt{a - 3} + 2} = \dfrac{\sqrt{a - 3} - 2}{\left( \sqrt{a - 3} + 2 \right)\left( \sqrt{a - 3} - 2 \right)} = \dfrac{\sqrt{a - 3} - 2}{a - 3 - 4} =\)
\(= \dfrac{\sqrt{a - 3} - 2}{a - 7}\)
5)
\(\dfrac{7}{3 - \sqrt{b + 2}} = \dfrac{7\left( 3 + \sqrt{b + 2} \right)}{\left( 3 - \sqrt{b + 2} \right)\left( 3 + \sqrt{b + 2} \right)} = \dfrac{7\left( 3 + \sqrt{b + 2} \right)}{9 - b - 2} =\)
\(= \dfrac{21 + 7\sqrt{b + 2}}{7 - b}\)
Ответ: 1) \(\frac{a}{2a - \sqrt{7b}} = \frac{2a^{2} + a\sqrt{7b}}{4a^{2} - 7b}\); 2) \(\frac{p - 3}{\sqrt{4 - p} - 1} = - \sqrt{4 - p} - 1\); 3) \(\frac{m}{m + \sqrt{n}} = \frac{m(m - \sqrt{n})}{m^{2} - n}\); 4) \(\frac{1}{\sqrt{a - 3} + 2} = \frac{\sqrt{a - 3} - 2}{a - 7}\); 5) \(\frac{7}{3 - \sqrt{b + 2}} = \frac{21 + 7\sqrt{b + 2}}{7 - b}\).
