Question

Thực hiện phép tính :

\(\frac{x+9}{x^2-9}-\frac{3}{x^2-3x}\)

Answer 1

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

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

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

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

Answer 2

\(\frac{x+9}{x^2-9}-\frac{3}{x^2-3x}\)

\(=\frac{x+9}{\left(x+3\right)\left(x-3\right)}-\frac{3}{x\left(x-3\right)}\)

\(=\frac{\left(x+9\right)x}{x\left(x+3\right)\left(x-3\right)}-\frac{3\left(x+3\right)}{x\left(x+3\right)\left(x-3\right)}\)

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

\(=\frac{\left(x+3\right)^2}{x\left(x+3\right)\left(x-3\right)}=\frac{x+3}{x\left(x-3\right)}\)

hok tốt ...

Answer 3

xl nha, mk quên -9 mà mk tưởng +9 , ^_^  , ko có hàng cuối nha

Answer 4

Diệp anh tú sai rồi : x² + 6x - 9 đâu phải hằng đẳng thức mà tách thành (x+3)² ?  phải là +9 mới đc nha

Answer 5

[(x+9)/(x^2-9)]-[3/(x^2-3x)]

={[(x+9)*x]/[(x-3)*(x+3)*x]} - {[3*(x+3)]/[x*(x-3)*(x+3)]}

={[x^2+9x]/[x*(x+3)*(x-3)]-{[3x+3]/[x*(x+3)*(x-3)]}

=(x^2+9x-3x-3)/x*(x+3)*(x-3)

=(x^2+6x-3)/x*(x+3)*(x-3)