Question

Bài 1 thực hiện phép tính

X+9  trên x² --  9       ---           3 trên x² + 3x

Bài 2 phân tích đa thức thành nhân tử

4x³  --   12x 

Answer 1

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

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

\(4x^3-12x=4x.\left(x^2-3x\right)\)

Answer 2

\(4x^3-12x=\left(4x\right)\left(x^2-3\right)\)

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

\(..........dễòi\)

Answer 3

Bài 1:

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

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

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

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

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

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

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