a: ĐKXĐ: x>=0; x<>1
\(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(x=33-8\sqrt2\)
\(=32-2\cdot4\sqrt2\cdot1+1\)
\(=\left(4\sqrt2-1\right)^2\)
Thay \(x=\left(4\sqrt2-1\right)^2\) vào A, ta được:
\(A=\frac{\sqrt{\left(4\sqrt2-1\right)^2}}{\left(4\sqrt2-1\right)^2+\sqrt{\left(4\sqrt2-1\right)^2}+1}=\frac{4\sqrt2-1}{33-8\sqrt2+4\sqrt2-1+1}\)
\(=\frac{4\sqrt2-1}{33-4\sqrt2}=\frac{\left(4\sqrt2-1\right)\left(33+4\sqrt2\right)}{\left(33-4\sqrt2\right)\left(33+4\sqrt2\right)}=\frac{132\sqrt2+32-33-4\sqrt2}{1057}=\frac{-1+128\sqrt2}{1057}\)
c: \(A<\frac13\)
=>\(\frac{\sqrt{x}}{x+\sqrt{x}+1}<\frac13\)
=>\(\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac13<0\)
=>\(\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}<0\)
=>\(-x+2\sqrt{x}-1<0\)
=>\(-\left(\sqrt{x}-1\right)^2<0\)
=>\(\left(\sqrt{x}-1\right)^2>0\) (Luôn đúng với mọi x thỏa mãn ĐKXĐ)
