Giải thích các bước giải:
Ta có:
$A=\dfrac{x}{x+5} -\dfrac{7x-15}{25-x^2}+\dfrac{3}{x-5}$
$\to A=\dfrac{x(x-5)}{(x+5)(x-5)}+\dfrac{7x-15}{x^2-25}+\dfrac{3(x+5)}{(x-5)(x+5)}$
$\to A=\dfrac{x^2-5x}{(x+5)(x-5)}+\dfrac{7x-15}{(x-5)(x+5)}+\dfrac{3x+15}{(x-5)(x+5)}$
$\to A=\dfrac{x^2-5x+7x-15+3x+15}{(x-5)(x+5)}$
$\to A=\dfrac{x^2+5x}{(x-5)(x+5)}$
$\to A=\dfrac{x(x+5)}{(x-5)(x+5)}$
$\to A=\dfrac{x}{x-5}$
Ta có:
$B=x^2-2x-15=(x^2-2x+1)-16=(x-1)^2-4^2=(x-1-4)(x-1+4)=(x-5)(x+3)$
$\to E=\dfrac{B}{A}=\dfrac{(x-5)^2(x+3)}{x}$
Để $|E-1|+1>E$
$\to |E-1|>E-1$
Mà $|E-1|\ge E-1,\quad\forall E$
Dấu = không xảy ra khi
$E-1\ne 0\to E\ne 1$
$\to \dfrac{(x-5)^2(x+3)}{x}\ne 1$
$\to (x-5)^2(x+3)\ne x$
$\to x^3-7x^2-6x+75\ne 0$
$\to x\ne \:5.81213\dots ,\:x\ne \:-3.04705\dots ,\:x\ne \:4.23492\dots $
