Question

A= (\(\dfrac{x+4}{3x+6}-\dfrac{1}{x^2+4x+4}\))(1+\(\dfrac{x-1}{x+5}\))

a/ Rút gọn

b/Tìm giá trị của x để A là một số nguyên

Answer 1

a: \(A=\left(\dfrac{x+4}{3\left(x+2\right)}-\dfrac{1}{\left(x+2\right)^2}\right)\cdot\dfrac{x+5+x-1}{x+5}\)

\(=\left(\dfrac{x^2+6x+8}{3\left(x+2\right)^2}-\dfrac{3}{3\left(x+2\right)^2}\right)\cdot\dfrac{2x+4}{x+5}\)

\(=\dfrac{\left(x+1\right)\left(x+5\right)}{3\left(x+2\right)^2}\cdot\dfrac{2\left(x+2\right)}{x+5}\)

\(=\dfrac{2\left(x+1\right)}{3\left(x+2\right)}=\dfrac{2x+2}{3x+6}\)

b: Để A là số nguyên thì \(6x+6⋮3x+6\)

\(\Leftrightarrow6x+12-6⋮3x+6\)

\(\Leftrightarrow3x+6\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{-\dfrac{5}{3};-\dfrac{7}{3};-\dfrac{4}{3};-\dfrac{8}{3};-1;-3;0;-4\right\}\)