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

Với \(-2< x< 2\Leftrightarrow\left\{{}\begin{matrix}x-2< 0\\x+2>0\end{matrix}\right.\Leftrightarrow\left(x-2\right)\left(x+2\right)< 0;x\ne-1\Leftrightarrow\left(x+1\right)^2>0\Leftrightarrow A< 0\)

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

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.

\(a,A=\dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4}\left(dkxd:x\ne\pm2\right)\)

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

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

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

Vậy \(A=\dfrac{\left(x+1\right)^2}{x^2-4}\)

\(b,\) Theo đề, ta có : \(-2< x< 2\) 

\(\Rightarrow x-2< 0;x+2>0;\left(x+1\right)^2>0\)

\(\Rightarrow A< 0\) hay phân thức luôn có giá trị âm

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

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

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

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

đầu bài sai rồi bạn ơi bạn cho x=0 thì \(A=\frac{3}{4}\)là số dương rồi 

\(a,Đkxđ:x\ne\pm2\)

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

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

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

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

b, Ta có: \(\left(x-2\right)\left(x+2\right)< 0;\forall-2< 2< 2;x\ne-1\)

Mà: \(\left(x+1\right)^2>0\left(\forall x\ne-1\right)\)

\(\Rightarrow\frac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}< 0;\forall-2< x< 2;x\ne-1\)

Vậy ............

1,

\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)

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

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)