Question

Bai 3: Cho P = (x ^ 2 - 4x - 4)/(4 - x ^ 2) + (3x + 9)/(x + 2) a. Rút gọn P b.Tính giá trị P khi x = 8 c.Tìm x để P là số nguyên

Answer 1

a ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(P=\dfrac{x^2-4x-4}{4-x^2}+\dfrac{3x+9}{x+2}\)

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

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

\(=\dfrac{-x^2+4x+4+3x^2-6x+9x-18}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x^2+7x-14}{\left(x-2\right)\left(x+2\right)}\)

b: khi x=8 thì \(P=\dfrac{2\cdot8^2+7\cdot8-14}{\left(8-2\right)\left(8+2\right)}=\dfrac{2\cdot64+56-14}{64-4}=\dfrac{17}{6}\)