x^8>=0; x^8-x^5>=0;

x^2>=0; x^2-x>=0;

x^8-x^5+X^2-X+1>=1;

x^8 - x^7 + x^2 - x + 1 

= x^7(x-1) + x(x-1) +1

= (x-1)(x^7 + x) + 1

= (x^2-x)(x^6+1) + 1

Ta có: x^2 - x lớn hơn hoặc = 0; x^6 + 1 >0

=> (x^2-x)(x^6+1) lơn hơn hoặc bằng 0

=> (x^2+1)(x^6+1) + 1 > 0

=> x^8 - x^7 + x^2 - x + 1 > 0 (đpcm)

\(x^8-x^7+x^6+x^5-x^4+x^3+x^2-x+1\)

\(=x^6\left(x^2-x+1\right)+x^3\left(x^2-x+1\right)+x^2-x+1\)

\(=\left(x^6+x^3+1\right)\left(x^2-x+1\right)\)

\(=\left[\left(x^3+\frac{1}{2}\right)^2+\frac{3}{4}\right]\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\) \(\forall x\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

a) \(D=\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}-1}{2}\)

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

b) Ta có: \(x+\sqrt{x}+1>1\)

Suy ra \(D=\frac{2}{x+\sqrt{x}+1}< 2\)

\(\Rightarrow0< D< 2\)