ko biết

câu này thi bn quy đòng bình thường mà tính thôi 

khai triển ra 

rồi tạo ra x= y để thay vào bạn cứ biến đổi 

như vậy thì sẽ ra thôi

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

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

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

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

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

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

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

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

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

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

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

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

hùi nãy mem nào k sai cho t T_T t buồn 

\(VT\ge6\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-2\left(xy+yz+zx\right)+2.\frac{9}{4\left(x+y+z\right)}\)

\(=6\left(x+y+z\right)^2-2.\frac{\left(x+y+z\right)^2}{3}+\frac{9}{2\left(x+y+z\right)}=6.\left(\frac{3}{4}\right)^2-2.\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{9}{2.\frac{3}{4}}\)

\(=\frac{27}{8}-\frac{3}{8}+6=9\)

\(\Rightarrow\)\(VT\ge9\) ( đpcm ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)

Chúc bạn học tốt ~ 

Ta có:

\(\left(y^2+y+1\right)\left(x^2+x+1\right)\)

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

\(=x^2y^2+x^2+y^2+2xy+2=x^2y^2+3\)

Ta lại có:

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

\(=\left(y-x\right)\left(x+y+1\right)=-2\left(x-y\right)\)

Theo đề bài ta có: (sửa đề luôn)

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

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

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

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

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

Em xin đóng góp cách 2 ạ 

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}\)

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

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

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

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

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

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

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

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

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

\(gt\Rightarrow y-1=-x\Rightarrow x-1=-y\)

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

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

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

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

\(\Leftrightarrow\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+2xy+x^2+y^2+2\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}=\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(dpcm\right)\)

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

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

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

Ta nhận thấy các số trong ngoặc đều dương.

=> Để A>0 thì y>0

Vậy để A>0 thì y>0 và với mọi x