Question

Biến đổi mỗi biểu thức sau thành một phân thức đại số :

a) \(\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}\)

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

Answer 1

Answer 2

Bài 46: (Sgk/57)

a) \(\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}\) = (1+\(\dfrac{1}{x}\)) : (1-\(\dfrac{1}{x}\)) = \(\dfrac{x+1}{x}:\dfrac{x-1}{x}=\dfrac{x+1}{x}.\dfrac{x}{x-1}=\dfrac{x+1}{x-1}\)

b)

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

Answer 3