Question

(\(\frac{1}{2}\)x+y)(......)=\(\frac{x^3+8y^3}{8}\)

 

Answer 1

Ta có \(\left(\frac{1}{2}x+y\right)\left(...\right)=\frac{x^3+8y^3}{8}\)

\(\Leftrightarrow8\left(\frac{1}{2}x+y\right)\left(...\right)=x^3-8y^3\)

\(\Leftrightarrow4\left(x+2y\right)\left(...\right)=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

\(\Rightarrow4\left(...\right)=x^2-2xy+4y^2\)

\(\Rightarrow\left(...\right)=\frac{x^2-2xy+4y^2}{4}\)

Vậy đccm

#Học tốt

Answer 2

Ta có VP = \(\frac{x^3+8y^3}{8}\)

VP=\(\frac{x^3}{8}+y^3\)=\(\left(\frac{x}{2}\right)^3+y^3\)=\(\left(\frac{x}{2}+y\right)\).\(\left(\frac{x^2}{4}-\frac{xy}{2}+y^2\right)\)

Vậy \(\left(\frac{x^2}{4}-\frac{xy}{2}+y^2\right)\)