\(a,Đkxđ:x\ne\pm2\)

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

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

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

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

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 ............

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

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

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

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

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

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

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

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

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

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

Vì x>-2 và x<2 nên (x-2)<0, x+2>0, \(\left(x+1\right)^2>0\). Suy ra A<0

a: \(\Delta=\left(2m-2\right)^2-4\left(-m-3\right)\)

\(=4m^2-8m+4+4m+12\)

\(=4m^2-4m+16\)

\(=\left(2m-1\right)^2+15>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

b: Theo đề, ta có:

\(\left(x_1+x_2\right)^2-2x_1x_2>=10\)

\(\Leftrightarrow\left(2m-2\right)^2-2\left(-m-3\right)>=10\)

\(\Leftrightarrow4m^2-8m+4+2m+6-10>=0\)

\(\Leftrightarrow4m^2-6m>=0\)

=>m<=0 hoặc m>=3/2