Question

Cho A= \(\dfrac{x+19}{x^2-9}\) + \(\dfrac{2}{x+3}\)

a) Rút gọn A

b) Tìm x để A = \(\dfrac{-1}{2}\)

c) Tìm x ∈ N để A có giá trị nguyên

Giúp e với ạ!!

Answer 1

a:

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

 \(A=\dfrac{x+19}{x^2-9}+\dfrac{2}{x+3}\)

\(=\dfrac{x+19}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{x+3}\)

\(=\dfrac{x+19+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+13}{\left(x-3\right)\left(x+3\right)}\)

b: \(A=-\dfrac{1}{2}\)

=>\(\dfrac{3x+13}{x^2-9}=\dfrac{-1}{2}\)

=>\(x^2-9=-2\left(3x+13\right)=-6x-26\)

=>\(x^2+6x-9+26=0\)

=>\(x^2+6x+17=0\)

=>\(\left(x+3\right)^2+8=0\)(vô lý)

=>\(x\in\varnothing\)