Question

công trừ phân thức

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

Answer 1

Toán lớp 8 hở mày ?

Answer 2

ĐKXĐ : \(x\ne-1\)

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

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

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

\(=\frac{3\left(x+1\right)}{2\left(x+1\right)\cdot3\left(x+1\right)}-\frac{2\left(x-1\right)}{3\left(x+1\right)^2\cdot2}\)

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

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

\(=\frac{x+5}{6\left(x+1\right)^2}\)