Đáp án: a.$P=\dfrac{2x+1}{x+3}$ b.$x>-3$ c.$P=\dfrac{17}{21}$
Giải thích các bước giải:
a.Ta có :
$A=\dfrac{x}{x-3}-\dfrac{x+1}{x+3}+\dfrac{3x+2}{9-x^2}$
$\to A=\dfrac{x}{x-3}-\dfrac{x+1}{x+3}+\dfrac{3x+2}{-\left(x+3\right)\left(x-3\right)}$
$\to A=\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{-\left(3x+2\right)}{\left(x-3\right)\left(x+3\right)}$
$\to A=\dfrac{x\left(x+3\right)-\left(x+1\right)\left(x-3\right)-\left(3x+2\right)}{\left(x-3\right)\left(x+3\right)}$
$\to A=\dfrac{x^2+3x-x^2+2x+3-3x-2}{\left(x-3\right)\left(x+3\right)}$$
$\to A=\dfrac{2x+1}{\left(x-3\right)\left(x+3\right)}$
Lại có:
$B=\dfrac1{x-3}$
$\to P=A:B=\dfrac{2x+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{1}{x-3}$
$\to P=\dfrac{2x+1}{\left(x-3\right)\left(x+3\right)}.(x-3)$
$\to P=\dfrac{2x+1}{x+3}$
b.Để $P<2$
$\to \dfrac{2x+1}{x+3}<2$
$\to \dfrac{2x+1}{x+3}-2<0$
$\to \dfrac{2x+1-2\left(x+3\right)}{x+3}<0$
$\to \dfrac{-5}{x+3}<0$
$\to x+3>0$
$\to x>-3$
c.Ta có :
$\dfrac{3x-1}{x-3}=B.|5-2x|$
$\to \dfrac{3x-1}{x-3}=\dfrac1{x-3}.|5-2x|$
$\to 3x-1=|5-2x|$
$+) x\le \dfrac52\to 3x-1=5-2x\to x=\dfrac65$ (chọn )
$+) x>\dfrac52\to 3x-1=-\left(5-2x\right)\to x=-4$ (loại vì $x>\dfrac52$)
$\to x=\dfrac65$
$\to P=\dfrac{2x+1}{x+3}=\dfrac{2\cdot\dfrac65+1}{\dfrac65+3}=\dfrac{17}{21}$
