$f(x)=\frac{3}{x^4-x^3+x-1}-\frac{1}{x^4+x^3-x-1}-\frac{4}{x^5-x^4+x^3-x^2+x-1}
=\frac{3}{(x-1)(x+1)(x^2-x+1)}-\frac{1}{(x+1)(x-1)(x^2+x+1)}-\frac{4}{(x-1)(x^4+x^2+1)}
=\frac{(3x^2+3x+3)-(x^2-x+1)}{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}-\frac{4}{(x-1)(x^4+x^2+1)}
=\frac{2(x+1)^2}{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}$-$\frac{4}{(x-1)(x^4+x^2+1)}$
$=\frac{2(x+1)}{(x-1)(x^2-x+1)(x^2+x+1)}-\frac{4}{(x-1)(x^4+x^2+1)}=
\frac{2(x+1)}{(x-1)(x^4+x^2+1)}-\frac{4}{(x-1) (x^4+x^2+1)}
=\frac{2x-2}{(x^4+x^2+1)(x-1)}
=\frac{2}{x^4+x^2+1}$
( chắc sai)
b) (nếu a đúng)
$\frac{2}{x^4+x^2+1}> 0 ()$
$\frac{32}{9}-\frac{2}{x^4+x^2+1}=\frac{32x^4+32x^2+32-18}{9(x^4+x^2+1)}
=\frac{32x^4+32x^2+14}{9(x^4+x^2+1)}=\frac{32(x+\frac{1}{2})^2+6}{9(x^4+x^2+1)}> 0$
=>dpcm
Bài viết đã được chỉnh sửa nội dung bởi nhathongthai123: 06-07-2012 - 17:03