a)  \(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)

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

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

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

b) Xét tử  ta có:  \(\left(x+1\right)^2\ge0\)         (1)

   Xét mấu ta có:  \(x^2\ge0\Rightarrow x^2+1\ge1>0\)  (2)

Từ (1) và (2) \(\Rightarrow\) Phân thức trên k âm với mọi x   

Ta có: \(P=\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}=\frac{\left(x^3+1\right)\left(x+1\right)}{x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)}\)

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

Vì \(\hept{\begin{cases}x^2+1\ge1>0\\x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\end{cases}}\)

Nên mẫu số luôn luôn khác 0

Do đó: \(P=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}\)

Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\x^2+1>0\end{cases}\left(\forall x\right)}\) nên \(P\ge0\left(\forall x\right)\)

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

Do \(\left(x^2+1\right)\left(x^2-x+1\right)\ne0\)do đó không cần điều kiện của x

Vậy \(P=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}\)

\(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\x^2+1>0\forall x\end{cases}\Rightarrow P\ge0\forall x}\)

Phải đề thế này không

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

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

b/ Ta có: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\x^2+1>0\end{cases}\Rightarrow a=\frac{\left(x+1\right)^2}{x^2+1}\ge0}\)với mọi x

cảm ơn bạn

chịu

Lời giải:
$H=(x^3-3x^2+3x-1)-(x^3+8)+3(x^2-16)$

$=x^3-3x^2+3x-1-x^3-8+3x^2-48$

$=(x^3-x^3)+(-3x^2+3x^2)+3x+(-1-8-48)$

$=3x-57=3.\frac{-1}{2}-57=\frac{-117}{2}$