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

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

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

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

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

\(C=0\Leftrightarrow\frac{\left(x+1\right)^2}{x^2+1}=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

\(P=\frac{x^3+2x^2-5x-6}{x^2+x-6}=\frac{\left(x-2\right)\left(x+3\right)\left(x+1\right)}{\left(x-2\right)\left(x+3\right)}\)

=> x khác 2; x khác -3

không chắc đâu nhé :))

\(ĐKXĐ:x\ne2;x\ne4\)

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

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

\(\Leftrightarrow\frac{x^2-7x+12+x^2-4x+4}{x^2-6x+8}=-1\)

\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)

\(\Leftrightarrow x=3;x=\frac{8}{3}\)

Vậy tập nghiệm của phương trình  là \(S=\left\{3;\frac{8}{3}\right\}\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+2008\)

\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+2008\)

Đặt \(x^2+10x+20=t\)

Khi đó phương trình tương đương với:

\(\left(t-4\right)\left(t+4\right)+2008=t^2-16+2008=t^2+1992\)

Không hiểu phân tích ra như thế nào ?????

ĐKXĐ: x khác -1

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

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

<=> 2x - 1 = 2(x + 1) - x^2 + x - 1

<=> 2x - 1 = 2x + 2 - x^2 + x - 1

<=> 2x - 1 = 3x + 1 - x^2

<=> 2x - 1 - 3x - 1 + x^2 = 0

<=> -x - 2 + x^2 = 0

<=> (x - 2)(x + 1) = 0

<=> x - 2 = 0 hoặc x + 1 = 0

<=> x = 2 (tm) hoặc x = -1 (ktm)

=> x = 2