化简根式
十年级习题与解答

本文为十年级学生提供根式化简习题及解答。

为简化根式表达式，需掌握以下根式运算法则与性质

示例

2) 同次根式乘法公式 \[ \Large{\color{red}{ \left( \sqrt[n]{x} \right) \cdot \left( \sqrt[n]{y} \right) = \sqrt[n]{x \cdot y}} } \] 更多示例请参考根式乘法运算。 3) 同次根式除法公式 \[ \Large{\color{red}{ \dfrac{\sqrt[n]{x}}{\sqrt[n]{y}} = \sqrt[n]{\dfrac{x}{y}} }} \]

更多示例请参考根式除法运算。

4) 仅可对同类根式进行加减运算

示例

5) 可通过以下方式消去根号（分母有理化）

1. 示例1： \[ \color{red} { \sqrt{x} \cdot \sqrt{x} = x } \]
2. 示例2：\[ \color{red} { \sqrt[3]{x} \cdot \sqrt[3]{x} \cdot \sqrt[3]{x} = x } \]
3. 示例3： \[ \color{red} { (a - \sqrt{b})(a + \sqrt{b}) = (a)^2 - (\sqrt{b})^2 = a^2 - b } \]

更多示例请参考根式分母有理化。

习题集

1. \( \sqrt{128} \cdot \sqrt{32} \)
2. \( \sqrt{2} \cdot \sqrt{6} + 3\sqrt{12} \)
3. \( \dfrac{3\sqrt{14} + 4\sqrt{63}}{3\sqrt{7}} \)
4. \( \sqrt[3]{32} \sqrt[3]{16} \)
5. \( \sqrt[3]{\dfrac{64}{7}} \)
6. \( \dfrac{\sqrt[3]{2x} - \sqrt[3]{54x}}{\sqrt[3]{2}} \)
7. \( \dfrac{\sqrt{x} + \sqrt{x + 1}}{\sqrt{x} - \sqrt{x + 1}} \)

习题解答

1. 
   将128和32写作质因数乘积形式：\( 128 = 2^7, 32 = 2^5 \)，可得 \[ \sqrt{128} \cdot \sqrt{32} = \sqrt{2^7} \cdot \sqrt{2^5} = \sqrt{2^7 \cdot 2^5} = \sqrt{2^{12}} = \sqrt{(2^6)^2} = |2^6| = 64 \]
2. 
   运用乘积法则得 \( \sqrt{2} \cdot \sqrt{6} = \sqrt{12} \) \[ \sqrt{2} \cdot \sqrt{6} + 3\sqrt{12} = \sqrt{2 \cdot 6} + 3\sqrt{12} = \sqrt{12} + 3\sqrt{12} = \sqrt{12}(1 + 3) = 4\sqrt{12} \] \[ = 4\sqrt{4 \cdot 3} = 4\sqrt{4}\sqrt{3} = 4 \cdot 2 \cdot \sqrt{3} = 8\sqrt{3} \]
3. 
   将14和63写作质因数乘积形式 \( 14 = 2 \times 7, 63 = 3^2 \times 7 \) 并代入 \[ \dfrac{3\sqrt{14} + 4\sqrt{63}}{3\sqrt{7}} = \dfrac{3\sqrt{2 \cdot 7} + 4\sqrt{3^2 \cdot 7}}{3\sqrt{7}} \] \[ = \dfrac{3\sqrt{2}\sqrt{7} + 4 \cdot 3\sqrt{7}}{3\sqrt{7}} \] \[ = \dfrac{\sqrt{7}(3\sqrt{2} + 12)}{3\sqrt{7}} \] \[ = \dfrac{3\sqrt{2} + 12}{3} = \sqrt{2} + 4 \]
4. 
   将32和16写作质因数乘积形式 \( 32 = 2^5, 16 = 2^4 \) 并代入 \[ \sqrt[3]{32} \sqrt[3]{16} = \sqrt[3]{2^5} \sqrt[3]{2^4} = \sqrt[3]{2^5 \cdot 2^4} = \sqrt[3]{2^9} = \sqrt[3]{(2^3)^3} = 2^3 = 8 \]
5. 
   将64写作质因数乘积形式 \( 64 = 2^6 \) 并代入 \[ \sqrt[3]{\dfrac{64}{7}} = \sqrt[3]{\dfrac{2^6}{7}} = \sqrt[3]{\dfrac{(2^2)^3}{7}} = \dfrac{4}{\sqrt[3]{7}} \] 分子分母同乘 \(\left(\sqrt[3]{7}\right)^2\) 进行分母有理化 \[ = \dfrac{4}{\sqrt[3]{7}} \cdot \dfrac{\left(\sqrt[3]{7}\right)^2}{\left(\sqrt[3]{7}\right)^2} = \dfrac{4\left(\sqrt[3]{7}\right)^2}{7} = \dfrac{4\sqrt[3]{49}}{7} \]
6. 
   将54写作质因数乘积形式 \( 54 = 2 \times 3^3 \) 并代入 \[ \dfrac{\sqrt[3]{2x} - \sqrt[3]{54x}}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{2x} - \sqrt[3]{2 \cdot 3^3 x}}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{2x} - \sqrt[3]{2x} \cdot \sqrt[3]{3^3}}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{2x} - 3\sqrt[3]{2x}}{\sqrt[3]{2}} \] \[ = \dfrac{-2\sqrt[3]{2x}}{\sqrt[3]{2}} = -2\sqrt[3]{\dfrac{2x}{2}} = -2\sqrt[3]{x} \]
7. 
   分子分母同乘分母的共轭表达式 \[ \dfrac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} - \sqrt{x+1}} = \dfrac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} - \sqrt{x+1}} \cdot \dfrac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+1}} \] 展开并化简 \[ = \dfrac{(\sqrt{x})^2 + (\sqrt{x+1})^2 + 2\sqrt{x}\sqrt{x+1}}{(\sqrt{x})^2 - (\sqrt{x+1})^2} \] \[ = \dfrac{x + (x + 1) + 2\sqrt{x}\sqrt{x+1}}{x - (x + 1)} \] \[ = \dfrac{2x + 1 + 2\sqrt{x(x+1)}}{-1} \] \[ = -2x - 1 - 2\sqrt{x(x+1)} \]

拓展习题与答案

1. \(\sqrt[3]{25} \cdot \sqrt[3]{125}\)
2. \((5\sqrt[3]{64})(-3\sqrt[3]{16})\)
3. \((7\sqrt{\dfrac{2}{5}})(2\sqrt{10})\)
4. \(\sqrt[3]{2\dfrac{10}{27}}\)
5. \((\sqrt{17x})(\sqrt{34x})\)
6. \(\sqrt{4y^2}\)
7. \(\sqrt{8y^4}\)
8. \(\sqrt{25+144}\)
9. \(\sqrt[2n]{x^{2n}}\)（\(n\) 为正整数）
10. \((\sqrt{x-2})(4\sqrt{x-2})\)
11. \(\dfrac{\sqrt[3]{27a^3b^5}}{\sqrt[3]{8a^6b^2}}\)
12. \(\dfrac{\sqrt{x}-\sqrt{x-2}}{\sqrt{x}+\sqrt{x-2}}\)

拓展习题解答

1. \(\sqrt[3]{25} \cdot \sqrt[3]{125} = \sqrt[3]{3125} = 5\sqrt[3]{25}\)
2. \((5\sqrt[3]{64})(-3\sqrt[3]{16}) = -15\sqrt[3]{1024} = -15\sqrt[3]{64 \cdot 16} = -15 \cdot 4 \cdot \sqrt[3]{16} = -60\sqrt[3]{16}\)
3. \((7\sqrt{\dfrac{2}{5}})(2\sqrt{10}) = 14\sqrt{\dfrac{20}{5}} = 14\sqrt{4} = 14 \cdot 2 = 28\)
4. \(\sqrt[3]{2\dfrac{10}{27}} = \sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3}\)
5. \((\sqrt{17x})(\sqrt{34x}) = \sqrt{578x^2} = x\sqrt{578} = x\sqrt{289 \cdot 2} = 17x\sqrt{2}\)
6. \(\sqrt{4y^2} = 2|y|\)
7. \(\sqrt{8y^4} = 2y^2\sqrt{2}\)
8. \(\sqrt{25+144} = \sqrt{169} = 13\)
9. \(\sqrt[2n]{x^{2n}} = |x|\)
10. \((\sqrt{x-2})(4\sqrt{x-2}) = 4(x-2)\)
11. \(\dfrac{\sqrt[3]{27a^3b^5}}{\sqrt[3]{8a^6b^2}} = \dfrac{3ab\sqrt[3]{b^2}}{2a^2b\sqrt[3]{1}} = \dfrac{3b}{2a} \)
12. \(\dfrac{\sqrt{x}-\sqrt{x-2}}{\sqrt{x}+\sqrt{x-2}} = \dfrac{(\sqrt{x}-\sqrt{x-2})^2}{x - (x-2)} = \dfrac{x + (x-2) - 2\sqrt{x(x-2)}}{2} = \dfrac{2x - 2 - 2\sqrt{x(x-2)}}{2} = x - 1 - \sqrt{x(x-2)}\)

相关链接

• 高中数学（10、11和12年级）- 免费习题与答案
• 初中数学（6、7、8、9年级）- 免费习题与答案
• 小学数学习题集（四、五年级） - 免费试题与解析
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