10y-50...

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October 2021 1 7 Report

Сократите дроби:

1) 5-20a^2/20a+10

2) 3x^2+8x-3/x^2+3x

3) y^3+7y^2-60y/10y-50

Answers & Comments

90misha90

Verified answer

1) $\frac{5-20a^2}{20a+10} = \frac{(-1)*(-5)+(-5)*4a^2}{10*2a+10*1} = \frac{(-5)*[(-1)+4a^2]}{10(2a+1)} = -\frac{5*[(2a)^2-(1)^2]}{5*2*(2a+1)} =$
$=-\frac{[2a-1]*[2a+1]}{2*(2a+1)} = -\frac{2a-1}{2} = \frac{1-2a}{2}= \frac{1}{2}- \frac{2a}{2}= \frac{1}{2}-a$
2) $\frac{3x^2+8x-3}{x^2+3x} = \frac{3x^2+9x-x-3}{x(x+3)}= \frac{3x(x+3)-(x+3)}{x(x+3)}= \frac{(3x-1)(x+3)}{x(x+3)}= \frac{3x-1}{x}$
3) $\frac{y^3+7y^2-60y}{10y-50} = \frac{y(y^2+7y-60)}{10(y-5)} = \frac{y(y^2-5y+12y-12*5)}{10(y-5)}=$
$=\frac{y[(y-5)(y+12)]}{10(y-5)} = \frac{y(y-5)(y+12)}{10(y-5)} = \frac{y(y+12)}{10}$

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