Question

[Nguyễn Bùi Đại Hiệp]

Cho x+y=1 và xy\(\ne\)0.CMR:\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^{2^{ }}y^2+3}\)

Answer 1

[Khôi Bùi]

Ta có : \(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x\left(x^3-1\right)-y\left(y^3-1\right)}{\left(x^3-1\right)\left(y^3-1\right)}\)

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

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

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

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

\(=\frac{\left(x-y\right)\left[x^2+y^2-\left(x+y\right)^2\right]}{x^3y^3+3xy}\)

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

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

Answer 2

[Nguyễn Thành Trương]

Kết hợp với giả thiết nêu ra ở đề bài, ta có vài biến đổi sau:

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

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

Mặt khác, ta lại có: \(\left(x^2+x+1\right)\left(y^2+y+1\right)=x^2y^2+xy^2+y^2+x^2y+xy+y+x^2+x+1\)

\(=x^2y^2+\left[x^2+xy\left(x+y\right)+xy+y^2\right]+\left(x+y\right)+1=x^2y^2+\left(x+y\right)^2+2=x^2y^2+3\)

Khi đó, trừ đẳng thức \(\left(1\right)\) cho đẳng thức \(\left(2\right)\) vế theo vế, ta được:

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

Vậy, \(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=-\frac{2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)