Question

tính gtbt

\(\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}\)vs x= -8 ;y= 6

   thanks các bn trc nha 

Answer 1

đặt A= \(\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}=\left(\frac{x}{2}\right)^3+3.\left(\frac{x}{2}\right)^2.\left(\frac{y}{3}\right)+3\left(\frac{x}{2}\right)\left(\frac{y}{3}\right)^2+\left(\frac{y}{3}\right)^3\)

= \(\left(\frac{x}{2}+\frac{y}{3}\right)^3\)

thay x=-8 vfa y=6 ta đucọ 

A= \(\left(-\frac{8}{2}+\frac{6}{3}\right)^3=\left(-4+2\right)^3=\left(-2\right)^3=-8\)