ĐKXĐ \(x\ne-2\)

A=\(\frac{-2x+4}{x+2}\)=\(\frac{-2\left(x+2\right)+8}{x+2}=-2+\frac{8}{x+2}\)

Để A nguyên thì \(\frac{8}{x+2}\)nguyên   =>x+2 thuộc uw8bạn tự giải nhé

       \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)

\(\Leftrightarrow\)\(\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c\ge\frac{a+b+c}{2}+\left(a+b+c\right)\)

\(\Leftrightarrow\)\(\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}\ge\frac{3}{2}\left(a+b+c\right)\)

\(\Leftrightarrow\)\(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\ge\frac{3}{2}\left(a+b+c\right)\)

\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\) (luôn đúng  BĐT Netbitt)

C/m:    \(VT=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

       \(=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1-3\)

      \(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)

     \(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

     \(=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

Ta có:   \(x+\frac{1}{x}\ge2\)   (x > 0)     (*)

     \(\Leftrightarrow\)\(\frac{x^2+1}{x}\ge\frac{2x}{x}\)

    \(\Leftrightarrow\) \(\frac{x^2-2x+1}{x}\ge0\) 

   \(\Leftrightarrow\)\(\frac{\left(x-1\right)^2}{x}\ge0\) luôn đúng 

Dấu "=" xảy ra   \(\Leftrightarrow\)\(x=1\) 

ÁP dụng   BĐT (*) ta có:   

    \(VT=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

  \(VT\ge\frac{1}{2}.9-3=\frac{3}{2}\) 

\(\Rightarrow\)đpcm

áp dụng bất đẳng thức CAUCHY SCHAWRZ DẠNG PHÂN THỨC

\(\frac{a^2}{a+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)

vết nhầm

:v tự viết tự trả lời lun kìa haha !#Huy$%^&*(Ơ}