НУ ПОМАГИТЕ ВЫ МНЕ УЖЕ! УЖЕ И 20 И 25 БАЛЛОВ СТАВИЛА А РЕШЕНИЕТАК И НЕТ! Выполните...

Объяснeние:

№39.3

$1) \frac{17-12x}{x} -\frac{10}{x} = \frac{17-12x-10}{x} =\frac{7-12x}{x} \\2)\frac{12p-1}{3p^{2} } -\frac{1+3p}{3p^{2} } =\frac{12p-1-1-3p}{3p^{2} } =\frac{9p-2}{3p^{2} }$ $3) \frac{6y-3}{5y} -\frac{y+2}{5y} =\frac{6y-3-y-2}{5y} =\frac{5y-5}{5y} =\frac{5(y-1)}{5y} =\frac{y-1}{y}$

$4) \frac{5b}{6} -\frac{3a-2b}{6} =\frac{5b-3a+2b}{6} =\frac{7b-3a}{6}$

$5) \frac{5y-3}{7y} -\frac{3y+2}{7y} =\frac{5y-3-3y-2}{7y} =\frac{2y-5}{7y}$

$6) \frac{11x-5}{14x} +\frac{3x-2}{14x} =\frac{11x-5+3x-2}{14x}=\frac{14x-7}{14x} =\frac{7(2x-1)}{14x} =\frac{2x-1}{2x}$

$7) \frac{7y-13}{10y}-\frac{2y+3}{10y}=\frac{7y-13-2y-3}{10y}=\frac{5y-16}{10y} \\8) \frac{8c+25}{6c} +\frac{5-2c}{6c} =\frac{8c+25+5-2c}{6c} =\frac{6c+30}{6c}$

№39.4

$1) \frac{3p-q}{5p}-\frac{2p+6q}{5p}+\frac{p-4q}{5p} =\frac{3p-q-2p-6q+p-4q}{5p} =\frac{2p-11q}{5p}$

$2) \frac{5c-2d}{4c} -\frac{3d}{4c} +\frac{d-5c}{4c} =\frac{5c-2d-3d+d-5c}{4c} =\frac{-4d}{4c} =-\frac{d}{c}$

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