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Câu 14: 1) Tính giá trị của biểu thức: a) $\sqrt{64} - 2\sqrt{81} + \sqrt{121}$ Ta có: $\sqrt{64} = 8$ $\sqrt{81} = 9$ $\sqrt{121} = 11$ Vậy: $\sqrt{64} - 2\sqrt{81} + \sqrt{121} = 8 - 2 \times 9 + 11 = 8 - 18 + 11 = 1$ b) $\frac{\sqrt{14} - \sqrt{7}}{\sqrt{2} - 1} - \sqrt{28} - \sqrt{(\sqrt{7} - 3)^2}$ Ta có: $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$ $\sqrt{(\sqrt{7} - 3)^2} = |\sqrt{7} - 3| = 3 - \sqrt{7}$ (vì $\sqrt{7} < 3$) Vậy: $\frac{\sqrt{14} - \sqrt{7}}{\sqrt{2} - 1} - \sqrt{28} - \sqrt{(\sqrt{7} - 3)^2} = \frac{\sqrt{14} - \sqrt{7}}{\sqrt{2} - 1} - 2\sqrt{7} - (3 - \sqrt{7})$ $= \frac{\sqrt{14} - \sqrt{7}}{\sqrt{2} - 1} - 2\sqrt{7} - 3 + \sqrt{7}$ $= \frac{\sqrt{14} - \sqrt{7}}{\sqrt{2} - 1} - \sqrt{7} - 3$ 2) Rút gọn các biểu thức sau: a) $A = \sqrt{x^2 - 6x + 9} - \sqrt{x^2} \quad (0 \leq x < 3)$ Ta có: $x^2 - 6x + 9 = (x - 3)^2$ Vì $0 \leq x < 3$, nên $x - 3 < 0$. Do đó: $\sqrt{(x - 3)^2} = |x - 3| = 3 - x$ $\sqrt{x^2} = x$ Vậy: $A = 3 - x - x = 3 - 2x$ b) $B = \left(\frac{3}{x - 9} + \frac{1}{\sqrt{x} + 3}\right) : \frac{2024\sqrt{x}}{x - 9} \quad (x > 0, x \neq 9)$ Ta có: $B = \left(\frac{3}{x - 9} + \frac{1}{\sqrt{x} + 3}\right) \times \frac{x - 9}{2024\sqrt{x}}$ $= \left(\frac{3(\sqrt{x} + 3) + (x - 9)}{(x - 9)(\sqrt{x} + 3)}\right) \times \frac{x - 9}{2024\sqrt{x}}$ $= \left(\frac{3\sqrt{x} + 9 + x - 9}{(x - 9)(\sqrt{x} + 3)}\right) \times \frac{x - 9}{2024\sqrt{x}}$ $= \left(\frac{x + 3\sqrt{x}}{(x - 9)(\sqrt{x} + 3)}\right) \times \frac{x - 9}{2024\sqrt{x}}$ $= \frac{x + 3\sqrt{x}}{(\sqrt{x} + 3)} \times \frac{1}{2024\sqrt{x}}$ $= \frac{\sqrt{x}(\sqrt{x} + 3)}{(\sqrt{x} + 3)} \times \frac{1}{2024\sqrt{x}}$ $= \frac{\sqrt{x}}{2024}$ Đáp số: 1) a) 1 b) $\frac{\sqrt{14} - \sqrt{7}}{\sqrt{2} - 1} - \sqrt{7} - 3$ 2) a) $3 - 2x$ b) $\frac{\sqrt{x}}{2024}$