a: Thay x=0 vào A, ta được:
\(A=\dfrac{0+1}{0-3}=\dfrac{1}{-3}=-\dfrac{1}{3}\)
b: \(B=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)+11x-3}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x-3}\)
c: \(P=\dfrac{B}{A}=\dfrac{3x}{x-3}:\dfrac{x+1}{x-3}=\dfrac{3x}{x+1}\)
Để P nguyên thì \(3x⋮x+1\)
=>\(3x+3-3⋮x+1\)
=>\(-3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
`#3107.101107`
`a.`
Tại `x = 0:`
`A = (0 + 1)/(0 - 3) = 1/(-3) = -1/3`
Vậy, `A = -1/3` tại `x = 0`
`b.`
`B= (2x)/(x+3) + (x+1)/(x-3) + (3 - 11x)/(9-x^2)`
\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{\left(3-x\right)\left(3+x\right)}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{\left(2x^2+x^2\right)+\left(11x-6x+4x\right)+\left(3-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{3x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{3x}{x-3}\)
`c.`
`P = B/A`
`=> P =` \(\dfrac{3x}{x-3}\div\dfrac{x+1}{x-3}\)
`P = (3x)/(x - 3) * (x - 3)/(x + 1)`
`P = (3x)/(x + 1)`
Ta có: `3x \vdots (x + 1)`
`=> (3x + 3 - 3) \vdots (x + 1)`
`=> -3 \vdots (x + 1)`
`=> x + 1 \in \text{Ư(-3)} = {+-1; +-3}`
`=>`\(\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=3\\x+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=2\\x=-4\end{matrix}\right.\)
