A=\(\frac{3-4x}{x^2+1}\)<=>\(A.\left(x^2+1\right)=3-4x\)

                      <=>\(Ax^2+4x+A-3=0\)

Để pt tồn tại x thì \(4-A.\left(A-3\right)=-A^2+3A+4>0\)

<=> \(A^2-3A-4< 0\)

<=>\(\left(A+1\right).\left(A-4\right)< 0\)

<=>\(-1< A< 4\)

Vậy min của A là: -1

       max của A là: 4

\(A=\frac{x}{3}+\frac{16}{3x}+\frac{15}{3}\ge2\sqrt{\frac{16x}{9x}}+\frac{15}{3}=\frac{23}{3}\)

Dấu ''='' xảy ra : <=> \(x=4\)

Vậy GTNN A = 23/3 <=> x = 4 

@phuongeieu : đề không cho x dương nên không thể xài Cauchy được nhé

\(G=\frac{2}{x^2+4y^2-4xy+4x-4y+10}\)

\(=\frac{2}{\left(x-2y\right)^2+2\left(x-2y\right)+1+2x+9}\)

\(=\frac{2}{\left(x-2y+1\right)^2+2x+9}\)

\(\frac{x^2+4x+4}{x^2-4}=\frac{x^2+3x+2}{A}\)

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

\(\frac{x^2+4x+4}{x^2-4}=\frac{x^2+3x+2}{A}\)

\(\Leftrightarrow\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x^2+x+2x+2}{A}\)

\(\Leftrightarrow\frac{x+2}{x-2}=\frac{x\left(x+1\right)+2\left(x+1\right)}{A}\Leftrightarrow\frac{x+2}{x-2}=\frac{\left(x+2\right)\left(x+1\right)}{A}\)

\(\Leftrightarrow A\left(x+2\right)=\left(x+2\right)\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow A=\left(x-2\right)\left(x+1\right)\)

Ta có: 

\(\left(x-y\right)^2\ge0\)

\(\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge x^2+y^2+2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2=1\)

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

Dấu \(=\)khi \(x=y=\frac{1}{2}\)