Краткое пояснение: Решаем задания на степени и упрощение выражений, используя свойства степеней и радикалов.
1. Вычислите:
-
6-3 ⋅ 65 = 6-3+5 = 62 = 36
-
45 : 48 = 45-8 = 4-3 = \(\frac{1}{4^3}\) = \(\frac{1}{64}\)
-
(52)-1 = 52⋅(-1) = 5-2 = \(\frac{1}{5^2}\) = \(\frac{1}{25}\)
-
4 ⋅ \((\frac{1}{4}\))-2 = 4 ⋅ 42 = 4 ⋅ 16 = 64
-
150 = 1
-
\(\sqrt{8}\) = \(\sqrt{2^3}\) = \(2^{\frac{3}{2}}\) = \(2^{1+\frac{1}{2}}\) = 2 ⋅ \(\sqrt{2}\)
-
\(\sqrt{81}\) = \(\sqrt{9^2}\) = 9
-
\(\sqrt[4]{\frac{1}{16}}\) = \(\sqrt[4]{\frac{1}{2^4}}\) = \(\frac{1}{2}\)
-
\(\sqrt[4]{81 ⋅ 16}\) = \(\sqrt[4]{81}\) ⋅ \(\sqrt[4]{16}\) = 3 ⋅ 2 = 6
-
\(\frac{\sqrt[3]{24}}{\sqrt[3]{3}}\) = \(\sqrt[3]{\frac{24}{3}}\) = \(\sqrt[3]{8}\) = 2
-
\(\sqrt{2} ⋅ \sqrt{8}\) = \(\sqrt{2 ⋅ 8}\) = \(\sqrt{16}\) = 4
-
\(\sqrt[5]{7^5}\) = 7
-
25\(\frac{1}{2}\) = \(\sqrt{25}\) = 5
-
-2 ⋅ 100\(\frac{1}{2}\) = -2 ⋅ \(\sqrt{100}\) = -2 ⋅ 10 = -20
-
64\(\frac{1}{3}\) = \(\sqrt[3]{64}\) = 4
2. Упростите выражение:
-
5-8 ⋅ 510 - 7-3 : 7-5 + \((\frac{3}{4}\))-2 = 52 - 72 + \((\frac{4}{3}\))2 = 25 - 49 + \(\frac{16}{9}\) = -24 + \(\frac{16}{9}\) = \(\frac{-216 + 16}{9}\) = \(\frac{-200}{9}\)
-
4\(\sqrt[4]{-27}\) - 0.3\(\sqrt{16}\) + \(\sqrt[8]{1}\) = 4\(\sqrt[4]{-27}\) - 0.3 ⋅ 4 + 1 = 4\(\sqrt[4]{-27}\) - 1.2 + 1 = 4\(\sqrt[4]{-27}\) - 0.2
-
(6.7 ⋅ 10-3) ⋅ (5 ⋅ 10-2) = 6.7 ⋅ 5 ⋅ 10-3 ⋅ 10-2 = 33.5 ⋅ 10-5 = 3.35 ⋅ 10-4
-
\(\frac{7^{-7} ⋅ 7^{-8}}{7^{-13}}\) = \(\frac{7^{-15}}{7^{-13}}\) = 7^{-15 - (-13)} = 7^{-2} = \(\frac{1}{7^2}\) = \(\frac{1}{49}\)
-
y\(\frac{2}{7}\) ⋅ y\(\frac{1}{6}\) = y\(\frac{2}{7} + \frac{1}{6}\) = y\(\frac{12+7}{42}\) = y\(\frac{19}{42}\)
-
\(\frac{x^{\frac{4}{9}} ⋅ x^{\frac{3}{5}}}{x^{\frac{5}{9}}}\) = \(\frac{x^{\frac{4}{9} + \frac{3}{5}}}{x^{\frac{5}{9}}}\) = \(\frac{x^{\frac{20+27}{45}}}{x^{\frac{5}{9}}}\) = \(\frac{x^{\frac{47}{45}}}{x^{\frac{25}{45}}}\) = x\(\frac{47}{45} - \frac{25}{45}\) = x\(\frac{22}{45}\)
-
(a8)\(\frac{7}{4}\) ⋅ a\(\frac{7}{2}\) = a8 ⋅ \(\frac{7}{4}\) ⋅ a\(\frac{7}{2}\) = a14 ⋅ a\(\frac{7}{2}\) = a14 + \(\frac{7}{2}\) = a\(\frac{28+7}{2}\) = a\(\frac{35}{2}\)
3. Представьте выражение в виде степени с основанием x:
\(\sqrt[4]{x} ⋅ \sqrt[3]{x^4}\) = x\(\frac{1}{4}\) ⋅ x\(\frac{4}{3}\) = x\(\frac{1}{4} + \frac{4}{3}\) = x\(\frac{3 + 16}{12}\) = x\(\frac{19}{12}\)
4. Сократите дробь:
-
\(\frac{a^3 + b^3}{a^3 - b^3}\) = \(\frac{(a+b)(a^2 - ab + b^2)}{(a-b)(a^2 + ab + b^2)}\)
-
\(\frac{6c^2 - c}{c^2 - 6}\) = \(\frac{c(6c - 1)}{c^2 - 6}\)
5. Упростите выражение:
-
((\(\frac{1}{a^4}\)) - b)2 - \(\sqrt{b}\) = \(\frac{1}{a^8}\) - \(\frac{2b}{a^4}\) + b2 - \(\sqrt{b}\)
-
\(\frac{\sqrt{(2 - 2\sqrt{2})^2} + \sqrt{(3 - 2\sqrt{2})^2}}{\sqrt{3} + \sqrt{5} ⋅ \sqrt{5-3}}\) = \(\frac{|2 - 2\sqrt{2}| + |3 - 2\sqrt{2}|}{\sqrt{3} + \sqrt{5} ⋅ \sqrt{2}}\) = \(\frac{2\sqrt{2} - 2 + 3 - 2\sqrt{2}}{\sqrt{3} + \sqrt{10}}\) = \(\frac{1}{\sqrt{3} + \sqrt{10}}\) = \(\frac{\sqrt{10} - \sqrt{3}}{7}\)
-
\(\frac{\sqrt{\sqrt{10} - 2} - \sqrt{\sqrt{10} + 2}}{\sqrt{24}}\) = \(\frac{\sqrt{\sqrt{10} - 2} - \sqrt{\sqrt{10} + 2}}{2\sqrt{6}}\) = \(\frac{\sqrt{6}}{12} ⋅ (\sqrt{\sqrt{10} + 2} - \sqrt{\sqrt{10} - 2})\)
-
\(\frac{(2a^2)^3 ⋅ (3b)^2}{(6a^3b)^2}\) = \(\frac{8a^6 ⋅ 9b^2}{36a^6b^2}\) = \(\frac{72a^6b^2}{36a^6b^2}\) = 2
6. Решите уравнение:
35x-3 = 27
35x-3 = 33
5x - 3 = 3
5x = 6
x = \(\frac{6}{5}\)
Ответ: x = 6/5