Giải thích các bước giải:
Câu 2:
a.Để biểu thức xác định
$\to\begin{cases} 2x+10\ne 0\\ x\ne 0\\ 2x(x+5)\ne 0\end{cases}$
$\to x\notin\{-5,0\}$
b.Ta có:
$P=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x(x+5)}$
$\to P=\dfrac{x^2+2x}{2(x+5)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x(x+5)}$
$\to P=\dfrac{\left(x^2+2x\right)x}{2\left(x+5\right)x}+\dfrac{\left(x-5\right)\cdot \:2\left(x+5\right)}{x\cdot \:2\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}$
$\to P=\dfrac{\left(x^2+2x\right)x+\left(x-5\right)\cdot \:2\left(x+5\right)+50-5x}{2x\left(x+5\right)}$
$\to P=\dfrac{x^3-5x+4x^2}{2x\left(x+5\right)}$
$\to P=\dfrac{x\left(x^2-5+4x\right)}{2x\left(x+5\right)}$
$\to P=\dfrac{x^2-5+4x}{2\left(x+5\right)}$
$\to P=\dfrac{\left(x-1\right)\left(x+5\right)}{2\left(x+5\right)}$
$\to P=\dfrac{x-1}{2}$
c.Để $P=-\dfrac14$
$\to \dfrac{x-1}{2}=-\dfrac14$
$\to x-1=-\dfrac12$
$\to x=\dfrac12$
d.Để $P>0\to \dfrac{x-1}{2}>0\to x-1>0\to x>1$
$P<0\to \dfrac{x-1}{2}<0\to x-1<0\to x<1$
Câu 2:
a.Để biểu thức xác định
$\to\begin{cases} x\ne 0\\ x+2\ne 0\end{cases}$
$\to x\notin\{0,-2\}$
b.Ta có:
$Q=\dfrac{(x+2)^2}{x}\cdot (1-\dfrac{x^2}{x+2})-\dfrac{x^2+6x+4}{x}$
$\to Q=\dfrac{(x+2)^2}{x}\cdot \dfrac{x+2-x^2}{x+2}-\dfrac{x^2+6x+4}{x}$
$\to Q=\dfrac{(x+2)(-x^2+x+2)}{x}-\dfrac{x^2+6x+4}{x}$
$\to Q=\dfrac{(x+2)(-x^2+x+2)-(x^2+6x+4)}{x}$
$\to Q=\dfrac{(-x^3-2x^2-2x)-(x^2+6x+4)}{x}$
$\to Q=\dfrac{-x^3-2x^2-2x}{x}$
$\to Q=-\dfrac{x\left(x^2+2x+2\right)}{x}$
$\to Q=-\left(x^2+2x+2\right)$
c.Ta có:
$Q=-\left(x^2+2x+2\right)$
$\to Q=-\left(x^2+2x+1+1\right)$
$\to Q=-\left((x+1)^2+1\right)$
$\to Q\le -(0+1)=-1<0$
$\to Q$ luôn âm
d.Từ câu c
$\to GTLN_Q=-1$ khi đó $x+1=0\to x=-1$
