(x-1)(x-2)(x+2)-(x-3)\(^3\)

=(x-1)(x\(^2\)-4)-(x-3)\(^3\)

(xy-1)(xy-2)-(xy-2)\(^2\)

=(xy-2)(xy-1-xy+2)

=xy-2

\(=\dfrac{\left(x-1\right)^3}{xy\left(x-1\right)-\left(x-1\right)}=\dfrac{\left(x-1\right)^3}{\left(xy-1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{xy-1}\left(xy\ne1;x\ne1\right)\)

a. \(\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}=\dfrac{x-2}{2x}\)

b. \(\dfrac{\dfrac{1}{2}-2}{2.\dfrac{1}{2}}=-1,5\)

a)2/x+4x+2/x^2-1

b)

B= \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).....\left(1-\dfrac{1}{20}\right)\)

B= \(\dfrac{1}{2}.\dfrac{2}{3}.....\dfrac{19}{20}\)

B= \(\dfrac{1.2.....19}{2.3.....20}\)

B= \(\dfrac{1}{20}\)

a) Ta có: \(C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)

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

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

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

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

b) Thay \(x=-\dfrac{3}{2}\) vào C, ta được:

\(C=\dfrac{-3}{2}:\left(\dfrac{9}{4}+1\right)=\dfrac{-3}{2}:\dfrac{13}{4}=\dfrac{-3}{2}\cdot\dfrac{4}{13}=\dfrac{-6}{13}\)

c) Ta có: \(C=\dfrac{1}{2}\)

nên \(\dfrac{x}{x^2+1}=\dfrac{1}{2}\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)(Loại)