Question

[Trần Thị Hà Phương]

\(\text{cho }xy\ne0\text{ và x + y = 1 }\)

\(\text{Chứng minh rằng}:\frac{x}{y^3-1}+\frac{y}{x^3-1}-\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)

Answer 1

[Lãnh Hạ Thiên Băng]

(chứng minh rằng\) x y 3 −1 - Online Math

Answer 2

[Tran Le Khanh Linh]

Ta có \(y^3-1=\left(y-1\right)\left(y^2+y+1\right)=-x\left(y^2+y+1\right)\)

(vì \(xy\ne0\Rightarrow x,y\ne0\))

\(\Rightarrow x-1\ne0;y-1\ne0\)

\(\Rightarrow\frac{x}{y^3-1}=\frac{-1}{y^2+y+1}\)

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

\(\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{-1}{y^2+y+1}+\frac{-1}{x^2+x+1}\)

\(=-\left(\frac{x^2+x+1+y^2+y+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\right)=-\left(\frac{\left(x+y\right)^2-2xy+\left(x+y\right)+2}{x^2y^2+\left(x+y\right)^2-2xy+xy\left(x+y\right)+xy+\left(x+y\right)+1}\right)\)

\(=-\frac{4-2xy}{x^2y^2+3}\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}-\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)