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)

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

\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)

a/ \(1-\dfrac{1}{1-\dfrac{1}{x}}=1-\dfrac{1}{\dfrac{x-1}{x}}=1-\dfrac{x}{x-1}=\dfrac{x-1-x}{x-1}=\dfrac{-1}{x-1}\)

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

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

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

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

\(=\dfrac{-1}{x-1}\)

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

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

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

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

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

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

b: \(=\dfrac{1-x^3+x^3}{1-x^3}=\dfrac{1}{1-x^3}\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{0;\pm1\right\}\\A=\dfrac{x+1}{x-1}\end{matrix}\right.\)