Question

Выполните действия:
а)$\frac{1}{x+1} - \frac{x-2}{3 x^{2} -3x}$
б)$\frac{x-2}{x+ x^{2} } + \frac{x}{ x^{2} -1}$
в)$\frac{2 x^{2} }{x*1} - 2x$

Answer 1

$\frac{1}{x+1}- \frac{2}{3x^2-3x}= \frac{3x(x-1)*1}{3x(x-1)*(x+1)}- \frac{2*(x+1)}{3x(x-1)*(x+1)}=$
$= \frac{3x(x-1)-2(x+1)}{3x(x-1)*(x+1)} = \frac{3x^2-3x-2x-2}{3x(x-1)(x+1)}= \frac{3x^2-5x-2}{3x(x-1)(x+1)}$

$\frac{x-2}{x+x^2}+ \frac{x}{x^2-1}= \frac{x-2}{x(x+1)}+ \frac{x}{(x-1)(x+1)}= \frac{(x-2)*(x-1)}{x(x+1)*(x-1)}+ \frac{x*x}{x*(x-1)(x+1)}=$
$= \frac{(x-2)(x-1)+x^2}{x(x+1)(x-1)} = \frac{x^2-3x+2+x^2}{x(x+1)(x-1)}= = \frac{2x^2-3x+2}{x(x+1)(x-1)}$

Если $x \neq 0$, то:
$\frac{2x^2}{x*1} -2x= \frac{2x^2}{x} - \frac{2x*x}{x} = \frac{2x^2-2x^2}{x} = \frac{0*x^2}{x} =0$
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$\frac{2x^2}{x-1} -2x= \frac{2x^2}{x-1} - \frac{2x*(x-1)}{x-1} = \frac{2x^2-2x(x-1)}{x} = \frac{2x^2-2x^2+2x}{x-1} =\frac{2x}{x-1}$
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$\frac{2x^2}{x+1} -2x= \frac{2x^2}{x+1} - \frac{2x*(x+1)}{x+1} = \frac{2x^2-2x(x+1)}{x+1} = \frac{2x^2-2x^2-2x}{x+1} =-\frac{2x}{x+1}$

Answer 2

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