thuc hien phep tinh :
\(\frac{1}{x-1}-\frac{x^3-x}{x^2+1}\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)\)
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Ta có :\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x}\)
\(=\frac{4\left(x-2\right)\left(4-x\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}+\frac{2\left(x+2\right)\left(4-x\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)\(+\frac{\left(5x-6\right)\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{24x-8x^2-32}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}+\frac{4x-2x^2+16}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)\(+\frac{5x^2+4x-12}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{24x-8x^2-32+4x-2x^2+16+5x^2+4x-12}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{32x-5x^2-28}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}...\frac{2018}{2016}.\frac{2019}{2017}\\ =\frac{3.4.5...2018.2019}{1.2.3...2016.2017}\\ =\frac{2018.2019}{2}=1009.2019\)