\(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=\left(1+\frac{1}{x}\right):\left(1-\frac{1}{x}\right)\)

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

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

\(B=\left(1+\frac{2}{x-1}\right):\left(1+\frac{2x}{x^2+1}\right)\)

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

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

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

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

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

Chúc bạn học tốt !!!

\(A=1+\frac{1}{x}\)

\(=\frac{x}{x}+\frac{1}{x}\)

\(=\frac{x+1}{x}\)

\(\frac{x+\frac{1}{y}}{y+\frac{1}{x}}=\frac{\frac{xy}{y}}{\frac{xy}{x}}=\frac{xy}{y}.\frac{x}{xy}=\frac{x}{y}\)

\(\frac{x+\frac{1}{y}}{y+\frac{1}{x}}=\left(x+\frac{1}{y}\right):\left(y+\frac{1}{x}\right)=\frac{xy+1}{y}:\frac{xy+1}{x}=\frac{\left(xy+1\right)\cdot x}{\left(xy+1\right)\cdot y}=\frac{x}{y}\).

\(\dfrac{x^2-1}{\left(x+1\right)\left(x-3\right)}\)

\(=\dfrac{x^2-1^2}{\left(x+1\right)\left(x-3\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-3\right)}\)

\(=\dfrac{x-1}{x-3}\)

Vậy đã biến đổi phân thức đó thành một phân thức bằng nó và có tử bằng với đa thức \(A=x-1\)

\(\dfrac{8x^2-8x+2}{\left(4x-2\right)\left(15-x\right)}=\dfrac{2\left(4x^2-4x+1\right)}{2\left(2x-1\right)\left(15-x\right)}\\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)\left(15-x\right)}=\dfrac{2x-1}{15-x}=\dfrac{1-2x}{x-15}\\ =\dfrac{A}{x-15}\)

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)

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