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

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

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)

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

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

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

a) Ta có A = 2 x + 1 x : 2 x − 1 x = 2 x + 1 2 x − 1  

b) Ta có  B = a + 2 a − 2 : a 2 + 4 a + 4 a 2 + 2 a + 4 = a + 2 a − 2 . a 2 + 2 a + 4 ( a + 2 ) 2 = a 2 + 2 a + 4 a 2 − 4