分母有理化 - 带解答的习题集
本文为十年级学生提供关于根式表达式分母有理化的习题及解答。
分母有理化是指将分母中的根号消除的过程
例题详解
以下恒等式可用于有理化分式表达式的分母。
- \(\quad \sqrt{x} \cdot \sqrt{x} = (\sqrt{x})^2 = x\)
- \(\quad \sqrt[3]{x} \cdot (\sqrt[3]{x})^2 = (\sqrt[3]{x})^3 = x\)
- \(\quad (\sqrt{x} - \sqrt{y}) (\sqrt{x} + \sqrt{y}) = (\sqrt{x})^2 - (\sqrt{y})^2 = x - y\)
- \(\quad (x - \sqrt{y}) (x + \sqrt{y}) = x^2 - (\sqrt{y})^2 = x^2 - y\)
例题
对以下表达式的分母进行有理化，并尽可能简化。
\[
\dfrac{1}{\sqrt{2} }
\]
解答
由于分母含有 \( \sqrt 2 \)，将分子和分母同乘以 \( \sqrt 2 \) 并简化：
\[
\dfrac{1}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{(\sqrt{2})^2} = \dfrac{\sqrt{2}}{2}
\]
\[
\dfrac{1}{\sqrt[3]{x}}
\]
解答
由于分母含有 \( \sqrt[3]{x} \)，将分子和分母同乘以 \( \left( \sqrt[3]{x} \right)^2 \) 并简化：
\[
\dfrac{1}{\sqrt[3]{x}} = \dfrac{1}{\sqrt[3]{x}} \cdot \dfrac{\left( \sqrt[3]{x} \right)^2}{\left( \sqrt[3]{x} \right)^2} = \dfrac{\sqrt[3]{x^2}}{x}
\]
\[
\dfrac{4}{\sqrt{3} - \sqrt{2}}
\]
解答
由于分母含有表达式 \( \sqrt{3} - \sqrt{2} \)，将分子和分母同乘以它的共轭式 \( \sqrt{3} + \sqrt{2} \) 得到：
\[
\dfrac{4}{\sqrt{3} - \sqrt{2}} = \dfrac{4}{\sqrt{3} - \sqrt{2}} \cdot \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}
\]
\[
= \dfrac{4(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}
\]
\[
= \dfrac{4(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^{2} - (\sqrt{2})^{2}}
\]
\[
= \dfrac{4(\sqrt{3} + \sqrt{2})}{3 - 2} = 4(\sqrt{3} + \sqrt{2})
\]
\[
\dfrac{5x^{2}}{\sqrt[3]{x^{2}}}
\]
解答
由于分母含有表达式 \( \sqrt[3]{x^2} \)，将分子和分母同乘以 \( \left( \sqrt[3]{x^2} \right)^2 \) 得到：
\[
\dfrac{5x^2}{\sqrt[3]{x^2}} = \dfrac{5x^2}{\sqrt[3]{x^2}} \cdot \dfrac{\left( \sqrt[3]{x^2} \right)^2}{\left( \sqrt[3]{x^2} \right)^2}
\]
\[
= \dfrac{5x^2 \sqrt[3]{x^4}}{\left( \sqrt[3]{x^2} \right)^3}
\]
简化并约去公因式：
\[
= \dfrac{5x^2 \sqrt[3]{x^4}}{x^2} = 5\sqrt[3]{x^4} = 5x\sqrt[3]{x}
\]
\[
5) \quad \dfrac{x^2}{y + \sqrt{x^2 + y^2}}
\]
解答
由于分母含有表达式 \( y + \sqrt{x^2 + y^2} \)，将分子和分母同乘以它的共轭式 \( y - \sqrt{x^2 + y^2} \) 得到：
\[
\dfrac{x^2}{y + \sqrt{x^2 + y^2}} = \dfrac{x^2}{y + \sqrt{x^2 + y^2}} \cdot \dfrac{y - \sqrt{x^2 + y^2}}{y - \sqrt{x^2 + y^2}}
\]
\[
= \dfrac{x^2(y - \sqrt{x^2 + y^2})}{(y)^2 - (\sqrt{x^2 + y^2})^2}
\]
\[
= \dfrac{x^2(y - \sqrt{x^2 + y^2})}{y^2 - (x^2 + y^2)}
\]
\[
= \dfrac{x^2(y - \sqrt{x^2 + y^2})}{-x^2}
\]
\[
= -y + \sqrt{x^2 + y^2}
\]
练习题
对以下表达式的分母进行有理化，并尽可能简化。
- \( \quad
\dfrac{10}{\sqrt{5}} \)
- \( \quad 2\sqrt{2}\sqrt{3} - \dfrac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}}
\)
- \( \quad \dfrac{7x^4}{\sqrt[3]{x^4}}
\)
- \( \quad \dfrac{-x^2}{y + \sqrt{x^2 + y^2}}
\)
练习题解答
分子和分母同乘以 \( \sqrt 5 \)：
\[
\frac{10}{\sqrt{5}} = \frac{10}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{(\sqrt{5})^2} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}
\]
分子和分母同乘以 \( \sqrt{2} - \sqrt{3} \)：
\[
2\sqrt{2}\sqrt{3} - \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}} = 2\sqrt{2}\sqrt{3} - \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}} \cdot \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}}
\]
并简化：
\[
= 2\sqrt{2}\sqrt{3} - \frac{(\sqrt{2}-\sqrt{3})^2}{(\sqrt{2})^2-(\sqrt{3})^2}
\]
\[
= 2\sqrt{2}\sqrt{3} - \frac{(\sqrt{2})^2 + (\sqrt{3})^2 - 2\sqrt{2}\sqrt{3}}{2-3}
\]
\[
= 2\sqrt{2}\sqrt{3} - \frac{2+3-2\sqrt{2}\sqrt{3}}{-1}
\]
\[
= 2\sqrt{2}\sqrt{3} + 5 - 2\sqrt{2}\sqrt{3}
\]
\[
= 5
\]
分子和分母同乘以 \( \left( \sqrt[3]{x^4} \right)^2 \)：
\[
\frac{7x^4}{\sqrt[3]{x^4}} = \frac{7x^4}{\sqrt[3]{x^4}} \cdot \frac{\left( \sqrt[3]{x^4} \right)^2}{\left( \sqrt[3]{x^4} \right)^2}
\]
并简化：
\[
= \frac{7x^4 \sqrt[3]{x^8}}{\left( \sqrt[3]{x^4} \right)^3} = \frac{7x^4 \sqrt[3]{x^8}}{x^4} = 7 \sqrt[3]{x^8} = 7x^2 \sqrt[3]{x^2}
\]
分子和分母同乘以 \( y - \sqrt{x^2 + y^2} \)：
\[
\frac{-x^2}{y + \sqrt{x^2 + y^2}} = \frac{-x^2}{y + \sqrt{x^2 + y^2}} \cdot \frac{y - \sqrt{x^2 + y^2}}{y - \sqrt{x^2 + y^2}}
\]
并简化：
\[
= \frac{-x^2(y - \sqrt{x^2 + y^2})}{y^2 - (x^2 + y^2)}
\]
\[
= \frac{-x^2(y - \sqrt{x^2 + y^2})}{-x^2} = y - \sqrt{x^2 + y^2}
\]
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