Question

Answer 1

Answer 2

a) Khi \(x=-1\)

\(A=\dfrac{2\left(-1\right)-1}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}\)

b)

\(B=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\\ B=\dfrac{3x+15+x-3-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{2x+6}{\left(x+3\right)\left(x-3\right)}\\ B=\dfrac{2}{x-3}\)

c)\(P=\dfrac{B}{A}=\dfrac{2}{x-3}:\dfrac{2x-1}{x-3}=\dfrac{2}{2x-1}\)

Để \(P< 0\)

\(\Leftrightarrow\dfrac{2}{2x-1}< 0\\ \Leftrightarrow2x-1< 0\left(2>0\right)\\ \Leftrightarrow x< \dfrac{1}{2}\)

Answer 3

a) Thay x=-1 vào A, ta được:

\(A=\dfrac{2\cdot\left(-1\right)-1}{-1-3}=\dfrac{-3}{-4}=\dfrac{3}{4}\)