Question
Simplify the expression
(1+\dfrac{x^2-1}{x^2-2x+1} )÷\dfrac{1}{x-1}
Answer
(1+\dfrac{x^2-1}{x^2-2x+1} )÷\dfrac{1}{x-1}
=[1+\dfrac{x-1)(x+1)}{(x-1)^2} ]\cdotp (x-1)
=( 1+x+1)(x-1)
=x^2+x-2
Simplify the expression
(1+\dfrac{x^2-1}{x^2-2x+1} )÷\dfrac{1}{x-1}
(1+\dfrac{x^2-1}{x^2-2x+1} )÷\dfrac{1}{x-1}
=[1+\dfrac{x-1)(x+1)}{(x-1)^2} ]\cdotp (x-1)
=( 1+x+1)(x-1)
=x^2+x-2