a) \(A=\dfrac{3x+15}{x^2+10x+25}\left(x\ne-5\right)\)
\(A=\dfrac{3\left(x+5\right)}{x^2+2.x.5+5^2}\)
\(A=\dfrac{3\left(x+5\right)}{\left(x+5\right)^2}\)
\(A=\dfrac{3}{x+5}\)
Khi \(x=10\) thì \(A=\dfrac{3}{10+5}=\dfrac{1}{5}\)
Vậy khi \(x=10\) thì \(A=\dfrac{1}{5}\)
b) \(B=\dfrac{x}{x+3}-\dfrac{2x}{3-x}-\dfrac{3x^2+9}{x^2-9}\left(x\ne\pm3\right)\)
\(B=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{x\left(x-3\right)+2x\left(x+3\right)-\left(3x^2-9\right)}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{x^2-3x+2x^2+6x-3x^2+9}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{3}{x-3}\)
c) Ta có:
\(P=\dfrac{A}{B}=\) \(\dfrac{\dfrac{3}{x+5}}{\dfrac{3}{x-3}}=\dfrac{3}{x+5}.\dfrac{x-3}{3}=\dfrac{x-3}{x+5}\)
Để P nhận giá trị nguyên thì \(x-3⋮x+5\)
\(\Rightarrow x+5\inƯ\left(x-3\right)\)
