Bài 3:

\(\frac{3n+1}{5n+2}\)

Ta có : (3n +1) * 5 =15n + 5

            (5n+2) *3 = 15n + 6

Mà :  15n + 6 - (15n + 5 ) =1 

       =>\(\frac{3n+1}{5n+2}\) tối giản ( ĐPCM)

ĐKXĐ: \(x\ne\pm2;x\ne0\)

\(A=\left[\frac{4x\left(x-2\right)}{x^2-4}-\frac{8x^2}{x^2-4}\right]:\left[\frac{x-1}{x\left(x-2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)}\right]\)

\(=\frac{-4x^2-8x}{x^2-4}:\frac{-x+3}{x\left(x-2\right)}\)

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

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

Vì \(4x^2\ge0\)với mọi x nên: 

để A > 0 thì x - 3 >0             <=>        x > 3

Điều kiện: X khác -1

\(A=\left(\frac{4}{x+1}-1\right):\frac{9+x^2}{x^2+2x+1}\)

\(=>A=\left(\frac{4}{x+1}-\frac{x+1}{x+1}\right):\frac{9+x^2}{\left(x+1\right)^2}\)

\(=>A=\frac{4-x-1}{x+1}\cdot\frac{\left(x+1\right)^2}{9+x^2}\)

\(=>A=\frac{\left(3-x\right)\cdot\left(x+1\right)}{9+x^2}\)

\(\left(\frac{4}{x+1}-1\right):\frac{9+x^2}{x^2+2x+1}\) ĐKXĐ : \(x\ne-1\)

\(\Leftrightarrow\left(\frac{4}{x+1}-\frac{x+1}{x+1}\right).\frac{x^2+2x+1}{9+x^2}\)

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

\(=\frac{x+1}{3+x}\)