Question

Câu 4: Quy đồng mẫu thức các phân thức: \(\dfrac{x+1}{x-x^2}\)và \(\dfrac{x+2}{2-4x+2x^2}\).

Answer 1

Ta có: x - x² = x( 1 - x)

2 - 4x + 2x² = 2( 1- 2x + x²) = 2(1 - x)²

⇒ MTC= 2x(1 - x)²

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

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