a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)

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

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

\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

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.

\(a,\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(l\right)\\x=-2\left(l\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\Leftrightarrow A\in\varnothing\\ b,\text{ý bạn là rút gọn A hả?}\\ A=\dfrac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x+4}{\left(x-2\right)\left(x+2\right)}\)

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

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

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

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

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

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

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

f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)

g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)

h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)

n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)

p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)

m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

a: \(A=\left(\dfrac{2x^2+2}{x^3-1}+\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+3}{x^3-x^2+3x-3}\right):\dfrac{1}{x-1}\)

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

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

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

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

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

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

b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)

=>\(x^2+x+1-x⋮x^2+x+1\)

=>\(x⋮x^2+x+1\)

=>\(x^2+x⋮x^2+x+1\)

=>\(x^2+x+1-1⋮x^2+x+1\)

=>\(-1⋮x^2+x+1\)

=>\(x^2+x+1\in\left\{1;-1\right\}\)

=>\(x^2+x+1=1\)

=>x²+x=0

=>x(x+1)=0

=>\(x\in\left\{0;-1\right\}\)

\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}x^2+x+1\ge0\\x^2+1\ne0\end{matrix}\right.\)

Ta có:

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

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

+) \(x^2+1\ge1>0\forall x\)

Vậy biểu thức luôn xác định với mọi x

2) ĐKXĐ: \(\left\{{}\begin{matrix}x^2-2x+3>0\\x^2-x+1\ge0\end{matrix}\right.\)

Ta có: 

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

\(=\left(x-1\right)^2+2\ge2>0\forall x\)

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

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

Vậy biểu thức luôn xác định với mọi x