Question

Cho A= \(\dfrac{4}{x-3}\)và B= \(\dfrac{4x}{x^2-9}\)- \(\dfrac{x-3}{x+3}\) 

a) cho x= 6 thì giá trị A

b) Rút gọn B

c) P= B - A tìm x thuộc Z để P nhận giá trị lớn nhất.

Answer 1

a: Khi x=6 thì \(A=\dfrac{4}{6-3}=\dfrac{4}{3}\)

b: \(B=\dfrac{4x}{x^2-9}-\dfrac{x-3}{x+3}\)(ĐKXĐ: \(x\notin\left\{3;-3\right\}\))

\(=\dfrac{4x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-3}{x+3}\)

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

\(=\dfrac{4x-x^2+6x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{-x^2+10x-9}{\left(x+3\right)\left(x-3\right)}\)