Giải thích các bước giải:
Bài 6:
1.Ta có:
$B=\dfrac{5+x}{5-x}:\dfrac{25x^2}{x^2-10x+25}\cdot (\dfrac{5}{5-x}-\dfrac{25}{x^3+125}\cdot \dfrac{5x-25-x^2}{x-5})$
$\to B=\dfrac{5+x}{5-x}\cdot \dfrac{x^2-10x+25}{25x^2}\cdot (\dfrac{5}{5-x}-\dfrac{25}{(x+5)(x^2-5x+25)}\cdot \dfrac{x^2-5x+25}{5-x})$
$\to B=\dfrac{5+x}{5-x}\cdot \dfrac{x^2-10x+25}{25x^2}\cdot (\dfrac{5(5+x)}{(5-x)(5+x)}-\dfrac{25}{(x+5)(5-x)})$
$\to B=\dfrac{5+x}{5-x}\cdot \dfrac{(5-x)^2}{25x^2}\cdot \dfrac{5(5+x)-25}{(5-x)(5+x)}$
$\to B= \dfrac{(5+x)(5-x)}{25x^2}\cdot \dfrac{5(5+x-5)}{(5-x)(5+x)}$
$\to B= \dfrac{(5+x)(5-x)}{25x^2}\cdot \dfrac{5x}{(5-x)(5+x)}$
$\to B= \dfrac1{5x}$
2.Để $B$ được xác định
$\to\begin{cases}5-x\ne 0\\ x^2-10x+25\ne 0\\ 25x^2\ne 0\\x^3+125\ne 0\end{cases}$
$\to\begin{cases}x\ne 5\\ (x-5)^2\ne 0\\ x\ne 0\\(x+5)(x^2-5x+25)\ne 0\end{cases}$
$\to\begin{cases}x\ne 5\\ x\ne 0\\x+5\ne 0\end{cases}$
$\to\begin{cases}x\ne 5\\ x\ne 0\\x\ne -5\end{cases}$
3a.Khi $x=-0.1$
$\to B=\dfrac1{5\cdot (-0.1)}=-2$
b.Để $B>0\to \dfrac1{5x}>0\to x>0$ vì $\dfrac15>0$
