Question

Answer 1

a: \(A=\dfrac{x}{x+5}-\dfrac{7x-15}{25-x^2}+\dfrac{3}{x-5}\)

\(=\dfrac{x}{x+5}+\dfrac{7x-15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3}{x-5}\)

\(=\dfrac{x\left(x-5\right)+7x-15+3\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{x^2-5x+7x-15+3x+15}{\left(x+5\right)\left(x-5\right)}=\dfrac{x^2+5x}{\left(x+5\right)\left(x-5\right)}=\dfrac{x}{x-5}\)

b: |2x-1|=9

=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=5\left(loại\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

Khi x=-4 thì \(A=\dfrac{-4}{-4-5}=\dfrac{-4}{-9}=\dfrac{4}{9}\)

c: \(P=A\cdot B=\dfrac{x}{x-5}\cdot\left(x^2-2x-15\right)\)

\(=\dfrac{x}{x-5}\cdot\left(x-5\right)\left(x+3\right)=x\left(x+3\right)\)

\(=x^2+3x\)

\(=x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{4}>=-\dfrac{9}{4}\)

Dấu = xảy ra khi x=-3/2