Question

công trừ phân thức

\(\frac{1}{x-3}-\frac{3}{2x+6}-\frac{x}{2x^2-12x+18}\)

Answer 1

\(\frac{1}{x-3}-\frac{3}{2x+6}-\frac{x}{2x^2-12x+18}\)

\(=\frac{1}{x-3}-\frac{3}{2\left(x+3\right)}-\frac{x}{2\left(x^2-6x+9\right)}\)

\(=\frac{1}{x-3}-\frac{3}{2\left(x+3\right)}-\frac{x}{2\left(x-3\right)^2}\)

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

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

\(=\frac{2x^2-18-3x^2+18x-27-x^2-3x}{2\left(x-3\right)^2\left(x+3\right)}\)

\(=\frac{-2x^2+15x-45}{2\left(x-3\right)^2\left(x+3\right)}\)