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Đáp án:
$1) A= \dfrac{2x}{x^2-3x}+\dfrac{2x}{x^2-4x+3}+\dfrac{x}{x-1}$
$= \dfrac{2x}{x.(x-3)}+\dfrac{2x}{x^2-x-3x+3}+\dfrac{x}{x-1}$
$= \dfrac{2x}{x.(x-3)}+\dfrac{2x}{(x-3).(x-1)}+\dfrac{x}{x-1}$
$= \dfrac{2x.(x-1)}{x.(x-3).(x-1)}+\dfrac{2x.x}{x.(x-3).(x-1)}+\dfrac{x^2.(x-3)}{x.(x-3).(x-1)}$
$= \dfrac{2x^2-2x+2x^2+x^3-3x^2}{x.(x-3).(x-1)}$
$= \dfrac{(2x^2+2x^2-3x^2)-2x+x^3}{x.(x-3).(x-1)}$
$= \dfrac{x^3-2x+1}{x.(x-3).(x-1)}$
~~~~~~~~~~~~~~~~~~~~~
$2) B= \dfrac{x}{x+2}+\dfrac{2}{x-2}-\dfrac{4x}{4-x^2}$
$= \dfrac{x.(x-2)}{x^2-4}+\dfrac{2.(x+2)}{x^2-4}+\dfrac{4x}{x^2-4}$
$= \dfrac{x^2-2x+2x+4+4x}{x^2-4}$
$= \dfrac{x^2+4x+4}{(x-2).(x+2)}$
$= \dfrac{(x+2)^2}{(x-2).(x+2)}$
$= \dfrac{x+2}{x-2}$
~~~~~~~~~~~~~~~~~~~~~
$3) C= \dfrac{1+x}{3-x}-\dfrac{1-2x}{3+x}-\dfrac{x.(1-x)}{9-x^2}$
$= \dfrac{(1+x).(3+x)}{9-x^2}-\dfrac{(1-2x).(3-x)}{9-x^2}-\dfrac{x-x^2}{9-x^2}$
$= \dfrac{3+x+3x+x^2-(3-x-6x+2x^2)-x+x^2}{9-x^2}$
$= \dfrac{3+4x+x^2-2x^2+7x-3-x+x^2}{9-x^2}$
$= \dfrac{10x}{9-x^2}$
~~~~~~~~~~~~~~~~~~~~~~~~~
$4) D= \dfrac{5}{2x^2+6x}-\dfrac{4-3x^2}{x^2-9}-3$
$= \dfrac{5}{2x.(x+3)}-\dfrac{4-3x^2}{(x-3).(x+3)}-3$
$= \dfrac{5.(x-3)}{2x.(x+3).(x-3)}-\dfrac{(4-3x^2).2x}{2x.(x+3).(x-3)}- \dfrac{3.2x.(x^2-9)}{2x.(x+3).(x-3)}$
$= \dfrac{5x-15-8x+6x^3-6x.(x^2-9)}{2x.(x^2-9)}$
$= \dfrac{6x^3-3x-15-6x^3+54x}{2x.(x^2-9)}$
$= \dfrac{51x-15}{2x.(x^2-9)}$
~~~~~~~~~~~~~~~~~~~~~~~~
$5) E= \dfrac{3x+2}{x^2-2x+1}-\dfrac{6}{x^2-1}-\dfrac{3x-2}{x^2+2x+1}$
$= \dfrac{3x+2}{(x-1)^2}-\dfrac{6}{(x-1).(x+1)}-\dfrac{3x-2}{(x+1)^2}$
$= \dfrac{(3x+2).(x^2+2x+1)}{(x-1)^2.(x+1)^2}-\dfrac{6.(x^2-1)}{(x-1)^2.(x+1)^2}-\dfrac{(3x-2).(x^2-2x+1)}{(x-1)^2.(x+1)^2}$
$=\dfrac{3x^3+8x^2+7x+2-6x^2+6-3x^3+8x^2-7x+2}{(x-1)^2.(x+1)^2}$
$= \dfrac{10x^2+10}{(x-1)^2.(x+1)^2}$
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