Giải thích các bước giải:
v.Ta có:
$\dfrac{x^4-2x^3}{2x^4-x^3}$
$=\dfrac{x^3(x-2)}{x^3(2x-1)}$
$=\dfrac{x-2}{2x-1}$
u.Ta có:
$\dfrac{x^7-x^4}{x^6-1}$
$=\dfrac{x^4(x^3-1)}{(x^3-1)(x^3+1)}$
$=\dfrac{x^4}{x^3+1}$
ư.Ta có:
$\dfrac{(x+2)^2-(x-2)^2}{16x}$
$=\dfrac{(x+2-(x-2))(x+2+(x-2))}{16x}$
$=\dfrac{4\cdot 2x}{16x}$
$=\dfrac{8x}{16x}$
$=\dfrac12$
x.Ta có:
$\dfrac{24.5x^2-0.5y^2}{3.5x^2-0.5xy}$
$=\dfrac{0.5(49x^2-y^2)}{0.5x(7x-y)}$
$=\dfrac{0.5(7x-y)(7x+y)}{0.5x(7x-y)}$
$=\dfrac{7x+y}{x}$
f.Ta có:
$\dfrac{9-(x+5)^2}{x^2+4x+4}$
$=\dfrac{3^2-(x+5)^2}{(x+2)^2}$
$=\dfrac{(3-(x+5))(3+(x+5))}{(x+2)^2}$
$=\dfrac{-(x+2)(x+8)}{(x+2)^2}$
$=-\dfrac{x+8}{x+2}$
h.Ta có:
$\dfrac{5x^3+5x}{x^4-1}$
$=\dfrac{5x(x^2+1)}{(x^2+1)(x^2-1)}$
$=\dfrac{5x}{x^2-1}$
