✏️ 本章练习题

答案：\((x + 2)(x - 3) = x^2 - 3x + 2x - 6 = x^2 - x - 6\)

答案：
a) \(3x^2 + 6x = 3x(x + 2)\)
b) \(x^2 - 4 = (x - 2)(x + 2)\)

答案：
a) \(\frac{x^5}{x^2} = x^{5-2} = x^3\)
b) \((y^3)^{\frac{1}{2}} = y^{\frac{3}{2}}\)

答案：\(\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\)

答案：\((2x - 1)(x^2 + 3x + 2) = 2x \cdot x^2 + 2x \cdot 3x + 2x \cdot 2 - 1 \cdot x^2 - 1 \cdot 3x - 1 \cdot 2 = 2x^3 + 6x^2 + 4x - x^2 - 3x - 2 = 2x^3 + 5x^2 + x - 2\)

答案：需要找两个数相乘得\(2 \times -3 = -6\)，相加得\(5\)，即\(6\)和\(-1\)。所以\(2x^2 + 5x - 3 = (2x - 1)(x + 3)\)

答案：\(y^{\frac{1}{2}} = (\frac{1}{9}x^2)^{\frac{1}{2}} = \frac{1}{3}x\)。所以\(k = \frac{1}{3}\)，\(n = 1\)

答案：
\(3\sqrt{8} = 3 \cdot 2\sqrt{2} = 6\sqrt{2}\)
\(2\sqrt{18} = 2 \cdot 3\sqrt{2} = 6\sqrt{2}\)
\(\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}\)
所以\(6\sqrt{2} - 6\sqrt{2} + 4\sqrt{2} = 4\sqrt{2}\)

答案：\(\frac{3}{2 + \sqrt{5}} \times \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{3(2 - \sqrt{5})}{4 - 5} = \frac{3(2 - \sqrt{5})}{-1} = -3(2 - \sqrt{5}) = -6 + 3\sqrt{5}\)

答案：试根法，试\(x = 2\)：\(8 + 12 - 8 - 12 = 0\)，所以\(x - 2\)是因式。
其他因式：\((x^3 + 3x^2 - 4x - 12) \div (x - 2) = x^2 + 5x + 6 = (x + 2)(x + 3)\)
所以完全因式分解：\((x - 2)(x + 2)(x + 3)\)

答案：
\(\frac{4x^3}{\sqrt{x}} + \frac{x^{\frac{5}{2}}}{\sqrt{x}} = 4x^3 \cdot x^{-\frac{1}{2}} + x^{\frac{5}{2}} \cdot x^{-\frac{1}{2}} = 4x^{3 - \frac{1}{2}} + x^{\frac{5}{2} - \frac{1}{2}} = 4x^{\frac{5}{2}} + x^2\)
所以\(a = \frac{5}{2}\)，\(b = 2\)

答案：
\(243 = 3^5\)，\(\sqrt{3} = 3^{\frac{1}{2}}\)，所以\(3^5 \cdot 3^{\frac{1}{2}} = 3^{5 + \frac{1}{2}} = 3^{\frac{11}{2}}\)
所以\(a = \frac{11}{2}\)

证明：
左边：\(\frac{(2 - \sqrt{x})^2}{\sqrt{x}} = \frac{4 - 4\sqrt{x} + x}{\sqrt{x}} = \frac{4}{\sqrt{x}} - \frac{4\sqrt{x}}{\sqrt{x}} + \frac{x}{\sqrt{x}} = 4x^{-\frac{1}{2}} - 4 + x^{\frac{1}{2}}\)
右边：\(4x^{-\frac{1}{2}} - 4 + x^{\frac{1}{2}}\)
左右相等，故得证。

答案：
先化简分母：\(\sqrt{75} - \sqrt{50} = \sqrt{25 \times 3} - \sqrt{25 \times 2} = 5\sqrt{3} - 5\sqrt{2} = 5(\sqrt{3} - \sqrt{2})\)
所以\(\frac{5}{5(\sqrt{3} - \sqrt{2})} = \frac{1}{\sqrt{3} - \sqrt{2}}\)
有理化：\(\frac{1}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2}\)
所以\(a = 3\)，\(b = 2\)

答案：\((\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b\)

证明：
每一项\(\frac{1}{\sqrt{n} + \sqrt{n+1}} = \frac{1}{\sqrt{n} + \sqrt{n+1}} \times \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}\)
所以原式 = \((\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \dots + (\sqrt{25} - \sqrt{24}) = \sqrt{25} - \sqrt{1} = 5 - 1 = 4\)

答案：\((\sqrt{11} - 5)(5 - \sqrt{11}) = \sqrt{11} \cdot 5 + (-5) \cdot 5 + \sqrt{11} \cdot (-\sqrt{11}) + (-5) \cdot (-\sqrt{11}) = 5\sqrt{11} - 25 - 11 + 5\sqrt{11} = 10\sqrt{11} - 36\)

答案：\(x - 64x^3 = x(1 - 64x^2) = x(1 - 8x)(1 + 8x)\)

答案：
\(27 = 3^3\)，所以\(27^{2x + 1} = (3^3)^{2x + 1} = 3^{3(2x + 1)} = 3^{6x + 3}\)
所以\(y = 6x + 3\)