Question

Tính phép tinh sau :

a) \(x-1-\dfrac{x^2-4}{x+1}\)

b) \(\dfrac{3x-1}{6x+2}-\dfrac{3x+1}{2-6x}-\dfrac{6x}{9x^2-1}\)

c)\(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\)

d) \(\dfrac{2x^2+1}{x^3+1}-\dfrac{x-1}{x^2-x-1}-\dfrac{1}{x+1}\)

Answer 1

a) \(x-1-\dfrac{x^2-4}{x+1}=\dfrac{\left(x-1\right)\left(x+1\right)-\left(x^2-4\right)}{x+1}=\dfrac{x^2-1-x^2+4}{x+1}=\dfrac{3}{x+1}\)

b) \(\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{\left(3x-1\right)^2+\left(3x+1\right)^2-12x}{2\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

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

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

Answer 2

c: \(=\dfrac{1}{x-2}-\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\)

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

\(=\dfrac{x^2+2x-x^2-6x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

d: Sửa đề: \(\dfrac{2x^2+1}{x^3+1}-\dfrac{x-1}{x^2-x+1}-\dfrac{1}{x+1}\)

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

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