Question

bài 10.cho biểu thức :              \(P=\left(\dfrac{x}{x-2}+\dfrac{1}{x^2-4}\right):\dfrac{x+1}{x+2}vớix\ne-1,x\ne\pm2.\)

a,rút gọn P.

b.tính giá trị của P khi x=\(\dfrac{1}{2}\).

 

Answer 1

\(P=\left(\dfrac{x}{x-2}+\dfrac{1}{x^2-4}\right):\dfrac{x+1}{x+2}\left(x\ne-1;x\ne\pm2\right)\)

a. \(=\left[\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{1}{x^2-4}\right].\dfrac{x+2}{x+1}\)

\(=\dfrac{x^2+2x+1}{x^2-4}.\dfrac{x+2}{x+1}\)

\(=\dfrac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}.\dfrac{x+2}{x+1}\)

\(=\dfrac{x+1}{x-2}\)

b. Thay x = 1/2 vào P, ta đc:

\(P=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-2}=\dfrac{\dfrac{3}{2}}{-\dfrac{3}{2}}=-1\)