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   

Mình làm tắt thôi nhé

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

Dễ thấy \(A\ge0\)

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

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

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

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

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

Ta có : \(\frac{\left(x-1\right)^2}{x^2-x+1}=\frac{\left(x-1\right)^2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge0\)

=> Đpcm

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

\(1.a,Q=\frac{x+3}{2x+1}-\frac{x-7}{2x+1}=\frac{x+3}{2x+1}+\frac{7-x}{2x+1}\)

            \(=\frac{x+3+7-x}{2x+1}=\frac{10}{2x+1}\)

\(b,\) Vì \(x\inℤ\Rightarrow\left(2x+1\right)\inℤ\)

Q nhận giá trị nguyên \(\Leftrightarrow\frac{10}{2x+1}\) nhận giá trị nguyên

                                \(\Leftrightarrow10⋮2x+1\)

                                \(\Leftrightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

Mà \(\left(2x+1\right):2\) dư 1 nên \(2x+1=\pm1;\pm5\)

\(\Rightarrow x=-1;0;-3;2\)

Vậy.......................

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

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

Ta có: \(\left(x+1\right)^2\ge0;\forall x\)

\(x^2+1>1\); \(\forall x\)

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

Vậy \(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{\left(x+1\right)^2}{x^2+1}\ge0;\forall x\)

a)  \(A= \dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4} \\ =\dfrac{1}{x-2}+\dfrac{1}{x-2}+\dfrac{x^2+1}{(x-2)(x+2)} \\= \dfrac{x+2+x-2+x^2+1}{(x-2)(x+2)} \\=\dfrac{x^2+2x+1}{x^2-4} \\ =\dfrac{(x+1)^2}{(x-2)(x+2)}\)

b) Với mọi \(x\) thỏa mãn \(-2<x<2\) và \(x \ne -1\) thì \(x-2\) đều có giá trị âm, mà \(\begin{cases}(x+1)^2≥0\\x+2>0\\\end{cases}\) \( \Rightarrow\) Biểu thức A luôn có giá trị âm.