Question

cho em xin bài 1 ạ

Answer 1

Bài 1: 

a: Thay x=4 vào B, ta được:

\(B=\dfrac{4-3}{2\cdot4-4^2}=\dfrac{1}{-8}=\dfrac{-1}{8}\)

b: \(A=\dfrac{-4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{x-2}+\dfrac{x-2}{x+2}\)

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

\(=\dfrac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x}{x-2}\)

c: \(P=A:B=\dfrac{-4x}{x-2}\cdot\dfrac{-x\left(x-2\right)}{x-3}=\dfrac{4x^2}{x-3}\)

Để P<0 thì x-3<0

hay x<3