a)  \(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)

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

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

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

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

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

\(=\frac{-1}{\sqrt{x}+1}\)

ĐKXĐ: \(x\ne1\)

b)  \(x=\frac{4}{9}\Rightarrow\sqrt{x}=\frac{2}{3}\Rightarrow\sqrt{x}+1=\frac{5}{3}\Rightarrow A=\frac{-1}{\frac{5}{3}}=-\frac{3}{5}\)

c)  \(\left|A\right|=\frac{1}{3}\Leftrightarrow\frac{1}{\sqrt{x}+1}=\frac{1}{3}\Leftrightarrow\sqrt{x}+1=3\)

                                                               \(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(TM\right)\)

Áp dụng BĐT Bunhiacopski ta có:

\(\frac{x}{x^3+y^2+z}=\frac{x\left(\frac{1}{x}+1+z\right)}{\left(x^3+y^2+z\right)\left(\frac{1}{x}+1+z\right)}\le\frac{1+x+xz}{\left(x+y+z\right)^2}=\frac{1+x+xz}{9}\)

Tương tự rồi cộng lại ta được:

\(T\le\frac{3+x+y+z+xy+yz+zx}{9}=\frac{6+xy+yz+zx}{9}\le\frac{6+\frac{\left(x+y+z\right)^2}{3}}{9}=1\)

Dấu "=" xảy ra tại \(x=y=z=1\)