Question

\(\frac{x+3}{x^2-4}.\frac{8-12x+6x^2-x^3}{9x+27}\)                   

rut gon bieu thuc

Answer 1

\(\frac{x+3}{x^2-4}.\frac{8-12x+6^2-x^3}{9x+27}\)

\(=\frac{x+3}{x^2-4}.\frac{-x^3+6x^2-4}{9x+27}\)

\(=\frac{\left(x+3\right)\left(-x^3+6x^2-4\right)}{\left(x^2-4\right)\left(9x+27\right)}\)

\(=\frac{\left(x+3\right)\left(-x^3+6x-4\right)}{9\left(x+3\right)\left(x^2-4\right)}\)

\(=\frac{-x^3+6x^2-4}{9\left(x^2-4\right)}\)

Mk ko chắc

Answer 2

(x+3 )/ (x-2)(x+2) . [(2-x)^3 / 9(x+3)]

= -(x-2)^2 / [(x+2).9]