Bài 5:

a: \(A=\frac{-3\left(x+1\right)}{x^2-x-6}\)

\(=\frac{-3\left(x+1\right)}{x^2-3x+2x-6}\)

\(=\frac{-3\left(x+1\right)}{\left(x-3\right)\left(x+2\right)}\)

\(x^2-4=0\)

=>(x-2)(x+2)=0

=>x=2(nhận) hoặc x=-2(loại)

Khi x=2 thì \(A=\frac{-3\cdot\left(2+1\right)}{\left(2-3\right)\left(2+2\right)}=\frac{-3\cdot3}{\left(-1\right)\cdot4}=\frac94\)

b: \(B=\frac{2x}{x+3}-\frac{x}{3-x}-\frac{3x^2+9}{x^2-9}\)

\(=\frac{2x}{x+3}+\frac{x}{x-3}-\frac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x\left(x-3\right)+x\left(x+3\right)-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\frac{2x^2-6x+x^2+3x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{-3x-9}{\left(x-3\right)\left(x+3\right)}=\frac{-3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-3}{x-3}\)

c: P=B:A

\(=-\frac{3}{x-3}:\frac{-3\left(x+1\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\frac{3}{x-3}\cdot\frac{\left(x-3\right)\left(x+2\right)}{3\left(x+1\right)}=\frac{x+2}{x+1}\)

Để P nguyên thì x+2⋮x+1

=>x+1+1⋮x+1

=>1⋮x+1

=>x+1∈{1;-1}

=>x∈{0;-2}

mà x là số tự nhiên

nên x=0

ĐKXĐ: \(x\notin\left\{0;-9\right\}\)

Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)

Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)

\(\Leftrightarrow9x^2+81x+180=0\)

\(\Leftrightarrow x^2+9x+20=0\)

\(\Leftrightarrow x^2+4x+5x+20=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-4;-5}

1: Ta có: \(A=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1}{x^2+5x+5}\)

\(=\dfrac{\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1}{x^2+5x+5}\)

\(=\dfrac{\left(x^2+5x+5\right)^2}{x^2+5x+5}\)

\(=x^2+5x+5\)

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