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3 years ago

13

Addshow all steps please

1 answer:

choli [55]3 years ago

3 0

$\bf \cfrac{3}{x-2}+\cfrac{4}{x-1}-\cfrac{1}{x}$ so, no breaks on the denominator thus, the LCD will be all three then.

$\bf \cfrac{3}{x-2}+\cfrac{4}{x-1}-\cfrac{1}{x}\implies \cfrac{[3x(x-1)]+[4x(x-2)]-[(x-2)(x-1)]}{x(x-2)(x-1)}
\\\\\\
\cfrac{[3x^2-3x]+[4x^2-8x]-[x^2-3x+2]}{x(x-2)(x-1)}
\\\\\\
\cfrac{3x^2-3x+4x^2-8x-x^2+3x-2}{x(x-2)(x-1)}
\\\\\\
\cfrac{6x^2-8x-2}{x(x-2)(x-1)}\implies \cfrac{2(x^2-4x-1)}{x(x-2)(x-1)}$

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