Question

1) thực hiên các phép tính: a) \(\dfrac{x}{x-4}+\dfrac{4-x}{16-x^2}\) b) \(\dfrac{4x^2+8x+4}{x^2-1}:\dfrac{1+x}{1-x}\)

Answer 1

a) \(\dfrac{x}{x-4}+\dfrac{4-x}{16-x^2}=\dfrac{x\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{4-x}{\left(x-4\right)\left(x+4\right)}\)

\(=\dfrac{x^2+4x-4+x}{\left(x-4\right)\left(x+4\right)}=\dfrac{x^2+5x-4}{\left(x-4\right)\left(x+4\right)}\)

b) \(\dfrac{4x^2+8x+4}{x^2-1}:\dfrac{1+x}{1-x}=\dfrac{4\left(x^2+2x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{1-x}{1+x}\)

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

Answer 2

a) x-1

b) 4