(1) 1c−1c+1c=1c−1c2+1c=1c−cc2+1=c2+1c3+c−c=c2+1c3\dfrac{1}{c-\dfrac{1}{c+\dfrac{1}{c}}} = \dfrac{1}{c-\dfrac{1}{\dfrac{c^2+1}{c}}} = \dfrac{1}{c-\dfrac{c}{c^2+1}} = \dfrac{c^2+1}{c^3+c-c} = \dfrac{c^2+1}{c^3}c−c+c111=c−cc2+111=c−c2+1c1=c3+c−cc2+1=c3c2+1 (2) 1−11−11−11−1x=1−11−11−xx−1=1−11−x−1x−1−x=1−11+x−1=1−1x=x−1x1-\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{x}}}} = 1-\dfrac{1}{1-\dfrac{1}{1-\dfrac{x}{x-1}}} = 1-\dfrac{1}{1-\dfrac{x-1}{x-1-x}} = 1-\dfrac{1}{1+x-1} = 1-\dfrac{1}{x} = \dfrac{x-1}{x}1−1−1−1−x1111=1−1−1−x−1x11=1−1−x−1−xx−11=1−1+x−11=1−x1=xx−1 (3) xx−x+1x+3−x−1x=xx−x(x+1)x2+3x−x+1=xx−x(x+1)x2+2x+1=xx−x(x+1)(x+1)2=xx−x(x+1)=x(x+1)x2+x−x=x+1x\dfrac{x}{x-\dfrac{x+1}{x+3-\dfrac{x-1}{x}}} = \dfrac{x}{x-\dfrac{x(x+1)}{x^2+3x-x+1}} = \dfrac{x}{x-\dfrac{x(x+1)}{x^2+2x+1}} = \dfrac{x}{x-\dfrac{x(x+1)}{(x+1)^2}} = \dfrac{x}{x-\dfrac{x}{(x+1)}} = \dfrac{x(x+1)}{x^2+x-x} = \dfrac{x+1}{x}x−x+3−xx−1x+1x=x−x2+3x−x+1x(x+1)x=x−x2+2x+1x(x+1)x=x−(x+1)2x(x+1)x=x−(x+1)xx=x2+x−xx(x+1)=xx+1