根式的乘法 - 十年级习题与解答

本文提供十年级关于根式表达式乘法的习题及其解答。

同次根式的乘法运算

同次根式的乘法公式如下：
\[ \Large{\color{red}{\sqrt[n]{x} \cdot \sqrt[n]{y} = \sqrt[n]{x \cdot y}}} \]

例题

  1. \(\sqrt{3} \cdot \sqrt{12} = \sqrt{3 \cdot 12} = \sqrt{36} = 6\)
  2. \(\sqrt[3]{27x^3} = \sqrt[3]{27} \cdot \sqrt[3]{x^3} = 3x\)
  3. \(\sqrt[4]{\frac{1}{12}} \cdot \sqrt[4]{3} \cdot \sqrt[4]{64} = \sqrt[4]{\frac{1}{12} \cdot 3 \cdot 64} = \sqrt[4]{16} = 2\)
  4. \(\color{black}{3\sqrt[10]{x^3} \cdot 5\sqrt[10]{x^5} \cdot \sqrt[10]{x^2} = (3 \cdot 5) \sqrt[10]{x^3 \cdot x^5 \cdot x^2} = 15 \sqrt[10]{x^{10}} = 15|x| = 15x}\) (假设 \(x \geq 0\))

习题集

运用上述乘法公式化简下列表达式：
  1. \( 4\sqrt{2} \cdot 7\sqrt{32} \)
  2. \( 6\sqrt{x} \cdot \frac{2}{3}\sqrt{x} \)
  3. \( 2\sqrt[3]{\frac{1}{32}} \cdot \sqrt[3]{128} \cdot \sqrt[3]{16} \)
  4. \( -8\sqrt[5]{x^2} \cdot 2\sqrt[5]{x^3} \)
  5. \( \sqrt{(x-3)} \cdot \sqrt{(x-3)} \)
  6. \( \sqrt[8]{x} \cdot 5\sqrt[8]{x^4} \cdot 2\sqrt[8]{x^3} \)

习题解答

  1. \( \quad 4\sqrt{2} \cdot 7\sqrt{32} = (4 \cdot 7)\sqrt{2 \cdot 32} = 28\sqrt{64} = 28 \cdot 8 = 224 \)
  2. \( \quad 6\sqrt{x} \cdot \frac{2}{3}\sqrt{x} = \left(6 \cdot \frac{2}{3}\right)\sqrt{x \cdot x} = 4\sqrt{x^2} = 4|x| = 4x \quad (\text{假设 } x \geq 0) \)
  3. \( \quad 2\sqrt[3]{\frac{1}{32}} \cdot \sqrt[3]{128} \cdot \sqrt[3]{16} = 2\sqrt[3]{\frac{1}{32} \cdot 128 \cdot 16} = 2\sqrt[3]{64} = 2 \cdot 4 = 8 \)
  4. \( \quad -8\sqrt[5]{x^2} \cdot 2\sqrt[5]{x^3} = (-8 \cdot 2)\sqrt[5]{x^2 \cdot x^3} = -16\sqrt[5]{x^5} = -16x \)
  5. \( \quad \sqrt{(x-3)} \cdot \sqrt{(x-3)} = \sqrt{(x-3)(x-3)} = \sqrt{(x-3)^2} = |x-3| = x-3 \) (注：由于\(\sqrt{(x-3)}\)是实数，因此\((x-3) \geq 0\)，故有\(|x-3| = x-3\))
  6. \( \quad \sqrt[8]{x} \cdot 5\sqrt[8]{x^4} \cdot 2\sqrt[8]{x^3} = (5 \cdot 2)\sqrt[8]{x \cdot x^4 \cdot x^3} = 10\sqrt[8]{x^8} = 10|x| = 10x \) (假设 \(x \geq 0\))

参考资料与链接