Đáp án: a.$ A=\dfrac{-6}{\left(x-2\right)\left(x^2-x+6\right)}$
b.$A\in\{\dfrac{16}{23},\dfrac{16}{45}\}$
c.$ x>2$
Giải thích các bước giải:
a.ĐKXĐ: $x\ne \pm2$
Ta có:
$A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x}{x+2}\right)$
$\to A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{\left(x-2\right)\left(x+2\right)+10-x}{x+2}$
$\to A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{x^2-x+6}{x+2}$
$\to A=\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{x^2-x+6}$
$\to A=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{x^2-x+6}$
$\to A=\dfrac{-6}{\left(x-2\right)\left(x^2-x+6\right)}$
b.Ta có $|x|=\dfrac12\to x=\pm\dfrac12$
Nếu $x=\dfrac12\to A=\dfrac{-6}{\left(\dfrac12-2\right)\left(\left(\dfrac12\right)^2-\dfrac12+6\right)}=\dfrac{16}{23}$
Nếu $x=-\dfrac12\to A=\dfrac{-6}{\left(-\dfrac12-2\right)\left(\left(-\dfrac12\right)^2-\left(-\dfrac12\right)+6\right)}=\dfrac{16}{45}$
c.Để $A<0$
$\to \dfrac{-6}{\left(x-2\right)\left(x^2-x+6\right)}<0$
$\to \left(x-2\right)\left(x^2-x+6\right)>0$
Do $x^2-x+6=\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0$
$\to x-2>0\to x>2$
