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

\(A=\left(\frac{1+2x}{2.\left(2+x\right)}-\frac{x}{3.\left(x-2\right)}+\frac{2x^2}{3.\left(4-x^2\right)}\right).\frac{24-12x}{6+13x}\)

        \(=\left[\frac{3.\left(1+2x\right)\left(2-x\right)-2x\left(x+2\right)+4x^2}{2.3.\left(x+2\right)\left(2-x\right)}\right].\frac{24-12x}{6+13x}\)

          \(=\frac{6+9x-6x^2-2x^2-4x+4x^2}{6.\left(4-x^2\right)}.\frac{24-12x}{6+13x}\)

             \(=\frac{6+5x-4x^2}{6.\left(4-x^2\right)}.\frac{12.\left(2-x\right)}{6+13x}\) \(=\frac{\left(6+5x-4x^2\right).2}{\left(x+2\right)\left(6+13x\right)}=\frac{12+10x-8x^2}{13x^2+32x+12}\)

 Câu a đơn giản

b)

 \(A=\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}=\frac{\left(x^4-x^3\right)-\left(x-1\right)}{\left(x^4+x^3+\frac{x^2}{4}\right)+\left(\frac{11}{4}x^2+2x+\frac{4}{11}\right)+1-\frac{4}{11}}\)

\(=\frac{\left(x-1\right)\left(x^3-1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)

\(=\frac{\left(x-1\right)^2\left(x^2+x+1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)

\(=\frac{\left(x-1\right)^2\left[\left(x^2+x+0,25\right)+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)

\(=\frac{\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)

Vì \(\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]>0\)và \(\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}>0\)

nên \(A>0\)hay A ko âm

Nhớ k nha !

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

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

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

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

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

  \(\dfrac{3x+2}{x^2-2x+1}-\dfrac{6}{x^2-1}-\dfrac{3x-2}{x^2+2x+1}\)

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

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

= \(\dfrac{3x^3+8x^2+7x+2}{\left(x^2-1\right)^2}-\dfrac{6x^2-6}{\left(x^2-1\right)^2}-\dfrac{3x^3-8x^2+7x-2}{\left(x^2-1\right)^2}\)

= \(\dfrac{10x^2+10}{\left(x^2-1\right)^2}\)

= \(\dfrac{10\left(x^2+1\right)}{\left(x^2-1\right)^2}\)