Страница 208 номер 609, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением
Упростить выражение и найти его значение: 1) \(\frac{m^{\frac{1}{2}}n^{\frac{1}{5}} + n^{\frac{1}{3}}}{m + m^{\frac{1}{2}}n^{\frac{2}{15}}}\) при \(m = 0{,}04\), \(n = 243\); 2) \(\frac{m^{\frac{1}{2}}n^{-2} - n^{-\frac{3}{2}}}{m^{\frac{3}{4}} - m^{\frac{1}{4}}n^{\frac{1}{2}}}\) при \(m = 81\), \(n = 0{,}1\); 3) \(\left(1 + \sqrt{\frac{a - x}{a + x}}\right) \cdot \left(1 - \sqrt{\frac{a - x}{a + x}}\right)\) при \(a = 5\), \(x = 4\); 4) \(\frac{a + \sqrt{a^2 - x^2}}{a - \sqrt{a^2 - x^2}} - \frac{a - \sqrt{a^2 - x^2}}{a + \sqrt{a^2 - x^2}}\) при \(a = 3\), \(x = \sqrt{5}\).
1) \(\frac{m^{\frac{1}{2}}n^{\frac{1}{5}} + n^{\frac{1}{3}}}{m + m^{\frac{1}{2}}n^{\frac{2}{15}}}\) при \(m = 0{,}04\), \(n = 243\)
\(\dfrac{m^{\frac{1}{2}}n^{\frac{1}{5}} + n^{\frac{1}{3}}}{m + m^{\frac{1}{2}}n^{\frac{2}{15}}} = \dfrac{n^{\frac{1}{5}}\left(m^{\frac{1}{2}} + n^{\frac{1}{3} - \frac{1}{5}}\right)}{m^{\frac{1}{2}}\left(m^{\frac{1}{2}} + n^{\frac{2}{15}}\right)} = \dfrac{n^{\frac{1}{5}}}{m^{\frac{1}{2}}},\)
\(\dfrac{n^{\frac{1}{5}}}{m^{\frac{1}{2}}} = \dfrac{\sqrt[5]{243}}{\sqrt{0{,}04}} = \dfrac{3}{0{,}2} = 15;\)
2) \(\frac{m^{\frac{1}{2}}n^{-2} - n^{-\frac{3}{2}}}{m^{\frac{3}{4}} - m^{\frac{1}{4}}n^{\frac{1}{2}}}\) при \(m = 81\), \(n = 0{,}1\)
\(\dfrac{m^{\frac{1}{2}}n^{-2} - n^{-\frac{3}{2}}}{m^{\frac{3}{4}} - m^{\frac{1}{4}}n^{\frac{1}{2}}} = \dfrac{n^{-2}\left(m^{\frac{1}{2}} - n^{-\frac{3}{2} + 2}\right)}{m^{\frac{1}{4}}\left(m^{\frac{1}{2}} - n^{\frac{1}{2}}\right)} = \dfrac{n^{-2}}{m^{\frac{1}{4}}},\)
\(\dfrac{n^{-2}}{m^{\frac{1}{4}}} = \dfrac{\left(\frac{1}{10}\right)^{-2}}{\sqrt[4]{81}} = \dfrac{100}{3} = 33\dfrac{1}{3};\)
3) \(\left(1 + \sqrt{\frac{a - x}{a + x}}\right) \cdot \left(1 - \sqrt{\frac{a - x}{a + x}}\right)\) при \(a = 5\), \(x = 4\)
\(\left(1 + \sqrt{\dfrac{a - x}{a + x}}\right) \cdot \left(1 - \sqrt{\dfrac{a - x}{a + x}}\right) =\)
\(= 1 - \dfrac{a - x}{a + x} = \dfrac{a + x - a + x}{a + x} = \dfrac{2x}{a + x},\)
\(\dfrac{2x}{a + x} = \dfrac{2 \cdot 4}{5 + 4} = \dfrac{8}{9};\)
4) \(\frac{a + \sqrt{a^{2} - x^{2}}}{a - \sqrt{a^{2} - x^{2}}} - \frac{a - \sqrt{a^{2} - x^{2}}}{a + \sqrt{a^{2} - x^{2}}}\) при \(a = 3\), \(x = \sqrt{5}\)
\(\dfrac{a + \sqrt{a^{2} - x^{2}}}{a - \sqrt{a^{2} - x^{2}}} - \dfrac{a - \sqrt{a^{2} - x^{2}}}{a + \sqrt{a^{2} - x^{2}}} =\)
\(= \dfrac{\left(a + \sqrt{a^{2} - x^{2}}\right)^{2} - \left(a - \sqrt{a^{2} - x^{2}}\right)^{2}}{a^{2} - a^{2} + x^{2}} = \dfrac{4a\sqrt{\left(a^{2} - x^{2}\right)}}{x^{2}},\)
\(\dfrac{4a\sqrt{\left(a^{2} - x^{2}\right)}}{x^{2}} = \dfrac{12\sqrt{9 - 5}}{5} = \dfrac{24}{5} = 4\dfrac{4}{5}.\)
