Cho các đa thức:

A = 27x³y⁶ – \(\frac{1}{8}\)y³;          B = 9x²y⁴ + \(\frac{3}{2}\)xy³ + \(\frac{1}{4}\)y²;   C = 3xy² – \(\frac{1}{2}\)y.

Chứng minh rằng A : B = C.

Lời giải

Ta có B . C = \(\left( {9{x^2}{y^4} + \frac{3}{2}x{y^3} + \frac{1}{4}{y^2}} \right).\left( {3x{y^2} - \frac{1}{2}y} \right)\)

= \(9{x^2}{y^4}.\left( {3x{y^2} - \frac{1}{2}y} \right) + \frac{3}{2}x{y^3}.\left( {3x{y^2} - \frac{1}{2}y} \right) + \frac{1}{4}{y^2}.\left( {3x{y^2} - \frac{1}{2}y} \right)\)

= 27x³y⁶ – \(\frac{9}{2}{x^2}{y^5}\) + \(\frac{9}{2}{x^2}{y^5}\) – \(\frac{3}{4}x{y^4}\) + \(\frac{3}{4}x{y^4}\) – \(\frac{1}{8}{y^3}\)

= 27x³y⁶ – \(\frac{1}{8}\)y³  = A.

Do đó, B . C = A. Suy ra A : B = C (đcpcm).