Question

алгебраиснские дроби:
упростите выражение:
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Answer 1

Ответ:

$\displaystyle a)\ \ \frac{x-9}{x^2-9}-\frac{3}{3x-x^2}=\frac{x-9}{(x-3)(x+3)}+\frac{3}{x(x-3)}=\\\\\\=\frac{(x-9)\cdot x+3\cdot (x+3)}{x\cdot (x-3)(x+3)}=\frac{x^2-6x+9}{x(x-3)(x+3)}=\frac{(x-3)^2}{x\cdot (x-3)(x+3)}=\\\\=\frac{x-3}{x\cdot (x+3)}$

$b)\ \ \displaystyle \frac{1}{x-2}-\frac{6x}{x^3-8}=\frac{1}{x-2}-\frac{6x}{(x-2)(x^2+2x+4)}=\frac{x^2+2x+4-6x}{(x-2)(x^2+2x+4)}=\\\\\\=\frac{x^2-4x+4}{(x-2)(x^2+2x+4)}=\frac{(x-2)^2}{(x-2)(x^2+2x+4)}=\frac{x-2}{x^2+2x+4}$

$c)\ \ \displaystyle \frac{3}{x}+\frac{21}{x^2-7x}-\frac{4-x}{x-7}=\frac{3}{x}+\frac{21}{x(x-7)}-\frac{4-x}{x-7}=\\\\\\=\frac{3(x-7)+21-x\, (4-x)}{x(x-7)}=\frac{3x-21+21-4x+x^2}{x(x-7)}=\frac{x^2-x}{x(x-7)}=\\\\\\=\frac{x(x-1)}{x(x-7)}=\frac{x-1}{x-7}$