Bài 6 : Rút gọn biêu thức a) A=(y)/(x)sqrt ((x^2)/(y^4)) với xgt 0,yneq 0 b) B=2a^2sqrt ((b^4)/(4a^2)),alt 0 C) C=5absqrt ((25a^2)/(b^6)),alt 0,bgt 0 d) D=(sqrt (x))/(sqrt (x)-1)-(2sqrt (x)-1)/(x-sqrt (x)) xgt 0,xneq 1

a) $A = \frac{y}{x}\sqrt{\frac{x^2}{y^4}} = \frac{y}{x} \cdot \frac{\sqrt{x^2}}{\sqrt{y^4}} = \frac{y}{x} \cdot \frac{|x|}{|y^2|} = \frac{y}{x} \cdot \frac{x}{y^2} = \frac{1}{y}$ (vì x > 0, y ≠ 0)<br /><br />b) $B = 2a^2\sqrt{\frac{b^4}{4a^2}} = 2a^2 \cdot \frac{\sqrt{b^4}}{\sqrt{4a^2}} = 2a^2 \cdot \frac{|b^2|}{|2a|} = 2a^2 \cdot \frac{b^2}{2|a|} = 2a^2 \cdot \frac{b^2}{-2a} = -ab^2$ (vì a < 0)<br /><br />c) $C = 5ab\sqrt{\frac{25a^2}{b^6}} = 5ab \cdot \frac{\sqrt{25a^2}}{\sqrt{b^6}} = 5ab \cdot \frac{5|a|}{|b^3|} = 5ab \cdot \frac{5(-a)}{b^3} = 5ab \cdot \frac{-5a}{b^3} = \frac{-25a^2}{b^2}$ (vì a < 0, b > 0)<br /><br />d) $D = \frac{\sqrt{x}}{\sqrt{x}-1} - \frac{2\sqrt{x}-1}{x-\sqrt{x}} = \frac{\sqrt{x}}{\sqrt{x}-1} - \frac{2\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)} = \frac{x - (2\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{x - 2\sqrt{x} + 1}{\sqrt{x}(\sqrt{x}-1)} = \frac{(\sqrt{x}-1)^2}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}-1}{\sqrt{x}}$ (vì x > 0, x ≠ 1)<br />