Bài 30 (Trang 19 SGK Toán 9, Tập 1):

Rút gọn các biểu thức:

$\mathrm{a}) \frac{\mathrm{y}}{\mathrm{x}}.\sqrt{\frac{{\mathrm{x}}^2}{{\mathrm{y}}^4}} \mathrm{với} \mathrm{x} > 0 \mathrm{và} \mathrm{y} \ne 0;$

$\mathrm{b}) {\mathrm{y}}^2.\sqrt{\frac{{\mathrm{x}}^4}{4{\mathrm{y}}^2}} \mathrm{với} \mathrm{y} < 0;$

$\mathrm{c}) 5\mathrm{xy}.\sqrt{\frac{25{\mathrm{x}}^2}{{\mathrm{y}}^6}} \mathrm{với} \mathrm{x} < 0 \mathrm{và} \mathrm{y} > 0;$

$d) 0,2x^3{y}^3\sqrt{\frac{16}{x^4{y}^8}} với x\ne 0, y\ne 0.$

Hướng dẫn giải:

a) $\mathrm{a}) \frac{\mathrm{y}}{\mathrm{x}}.\sqrt{\frac{{\mathrm{x}}^2}{{\mathrm{y}}^4}} = \frac{\mathrm{y}}{\mathrm{x}}.\frac{\left|\mathrm{x}\right|}{{\mathrm{y}}^2} = \frac{\mathrm{y}}{\mathrm{x}}.\frac{\mathrm{x}}{{\mathrm{y}}^2} = \frac{1}{\mathrm{y}}.$

Vì x > 0 nên |x| = x : y² > 0 với mọi $y \ne 0$.

$\mathrm{b}) 2{\mathrm{y}}^2.\sqrt{\frac{{\mathrm{x}}^4}{4{\mathrm{y}}^2}} = 2{\mathrm{y}}^2.\frac{\left|\mathrm{x}\right|}{2\mathrm{y}} = 2{\mathrm{y}}^2.\frac{{\mathrm{x}}^2}{-2\mathrm{y}} = -{\mathrm{x}}^2\mathrm{y}$

Vì $x^2 \ge 0 với mọi x, y < 0 nên \left|2y\right| = -2y$

$\mathrm{c}) 5\mathrm{xy}.\sqrt{\frac{25{\mathrm{x}}^2}{{\mathrm{x}}^4{\mathrm{y}}^8}} = 5\mathrm{xy}\frac{\left|5\mathrm{x}\right|}{{\mathrm{y}}^3} = 5\mathrm{xy}\frac{-5\mathrm{x}}{{\mathrm{y}}^3} = \frac{25{\mathrm{x}}^2}{{\mathrm{y}}^2}$

Vì x < 0 nên |5x| = -5x:y > 0 nên |y³| = y³.

$d) 0,2x^3{y}^3.\sqrt{\frac{16}{x^4{y}^8}} = 0,2x^3{y}^3.\frac{\sqrt{16}}{\sqrt{{\left(x^2{y}^4\right)}^2}} = 0,2x^3{y}^3.\frac{4}{\left|x^2{y}^4\right|} = \frac{0,8x}{y}$

$\left(Vì x^2{y}^4 = {\left(x{y}^2\right)}^2 > 0 với mọi x\ne 0, y\ne 0\right)$