Question

эта сумативка по придмету алгебра ​

Answer 1

Решение:
$1.a) \frac{15a(a - b)}{40b(a - b) } = \frac{3a}{8b} \\ b) \frac{y ^{2} - y }{y {}^{2} } = \frac{y(y - 1)}{y {}^{2} } = \frac{y - 1}{y} \\ 2.a) \frac{12x - 7}{15x} + \frac{3x - 2}{15x} = \frac{12x + 3x - 7 - 2}{15x} = \frac{15x - 9}{15x} = \frac{3(5x - 3)}{15x} = \frac{5x - 3}{5x} \\ b) \frac{ax + ay}{xy {}^{2} } \times \frac{x {}^{2} y}{3x + 3y} = \frac{a(x + y) \times x {}^{2} y}{xy {}^{2} \times 3(x + y) } = \frac{ax}{3y} \\ 3.y = 70 \\ \frac{y {}^{2} - 6y + 9 }{y {}^{2} - 9 } \div \frac{10y - 30}{y {}^{2} + 3y } = \frac{(y - 3) {}^{2} }{(y - 3)(y + 3)} \times \frac{y(y + 3)}{10(y - 3)} = \frac{y}{10} \\ \frac{70}{10} = 7$
Объяснение:
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