Question

помогите,я прошу вас
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Answer 1

$1)\ \ \dfrac{17-12x}{x}-\dfrac{10}{x}=\dfrac{17-12x-10}{y}=\dfrac{7-12x}{x}\\\\\\2)\ \ \dfrac{12p-1}{3p^2}-\dfrac{1+3p}{3p^2}=\dfrac{12p-1-1-3p}{3p^2}=\dfrac{9p-2}{3p^2}\\\\\\3)\ \ \dfrac{6y-3}{5y}-\dfrac{y+2}{5y}=\dfrac{6y-3-y-2}{5y}=\dfrac{5y-5}{5y}=\dfrac{5(y-1)}{5y}=\dfrac{y-1}{y}\\\\\\4)\ \ \dfrac{5b}{6}-\dfrac{3a-2b}{6}=\dfrac{5b-3a+2b}{6}=\dfrac{7b-3a}{6}\\\\\\5)\ \ \dfrac{5y-3}{7y}-\dfrac{3y+2}{7y}=\dfrac{5y-3-3y-2}{7y}=\dfrac{2y-5}{7y}$

$6)\ \ \dfrac{11x-5}{14x}+\dfrac{3x-2}{14x}=\dfrac{11x-5+3x-2}{14x}=\dfrac{14x-7}{14x}=\dfrac{7(2x-1)}{14x}=\dfrac{2x-1}{2x}\\\\\\7)\ \ \dfrac{7y-13}{10y}-\dfrac{2y+3}{10y}=\dfrac{7y-13-2y-3}{10y}=\dfrac{5y-16}{10y}\\\\\\8)\ \ \dfrac{8c+25}{6c}+\dfrac{5-2c}{6c}=\dfrac{8c+25+5-2c}{6c}=\dfrac{6c+30}{6c}=\dfrac{6(c+5)}{6c}=\dfrac{c+5}{c}$