Giải thích các bước giải:
a.Ta có: $x^2-8x+7=0\to (x-1)(x-7)=0\to x=7$ vì $x\ne 1$
$\to B=\dfrac1{7-1}=\dfrac16$
b.Ta có:
$A=\dfrac{x^2+2}{x^3-1} +\dfrac{x+1}{x^2+x+1}$
$\to A=\dfrac{x^2+2}{(x-1)(x^2+x+1)} +\dfrac{(x+1)(x-1)}{(x-1)(x^2+x+1)}$
$\to A=\dfrac{x^2+2+(x+1)(x-1)}{(x-1)(x^2+x+1)}$
$\to A=\dfrac{x^2+2+x^2-1}{(x-1)(x^2+x+1)}$
$\to A=\dfrac{2x^2+1}{(x-1)(x^2+x+1)}$
$\to A=\dfrac{2x^2+1}{x^3-1}$
c.Ta có:
$S=A-B$
$\to S=\dfrac{2x^2+1}{(x-1)(x^2+x+1)}-\dfrac{x^2+x+1}{(x-1)(x^2+x+1)}$
$\to S=\dfrac{2x^2+1-(x^2+x+1)}{(x-1)(x^2+x+1)}$
$\to S=\dfrac{x^2-x}{(x-1)(x^2+x+1)}$
$\to S=\dfrac{x(x-1)}{(x-1)(x^2+x+1)}$
$\to S=\dfrac{x}{x^2+x+1}$
d.Để $S=\dfrac13$
$\to \dfrac{x}{x^2+x+1}=\dfrac13$
$\to x^2+x+1=3x$
$\to x^2-2x+1=0$
$\to (x-1)^2=0$
$\to x-1=0$
$\to x=1$
e.Ta có;
$S-\dfrac13=\dfrac{x}{x^2+x+1}-\dfrac13=\dfrac{3x-(x^2+x+1)}{3(x^2+x+1)}=\dfrac{-(x-1)^2}{3(x^2+x+1)}$
Mà $x^2+x+1=(x+\dfrac12)^2+\dfrac34>0\to \dfrac{-(x-1)^2}{3(x^2+x+1)}\le 0$
$\to S-\dfrac13\le 0$
$\to S\le\dfrac13$
