Задание 3 Параграф 14 ГДЗ Рабочая тетрадь 1 Мерзляк Полонский 7 класс (Алгебра)

$$(0{,}3x-0{,}6y)(0{,}3x+0{,}6y)+0{,}36y^2 = (0{,}3x)^2-(0{,}6y)^2+0{,}36y^2$$

$$= 0{,}09x^2-0{,}36y^2+0{,}36y^2 = 0{,}09x^2$$

$$\frac{9}{16}m^2-\left(\frac{3}{4}m+\frac{1}{7}n\right)\left(\frac{3}{4}m-\frac{1}{7}n\right) = \frac{9}{16}m^2-\left(\left(\frac{3}{4}m\right)^2-\left(\frac{1}{7}n\right)^2\right)$$

$$= \frac{9}{16}m^2-\left(\frac{9}{16}m^2-\frac{1}{49}n^2\right) = \frac{9}{16}m^2-\frac{9}{16}m^2+\frac{1}{49}n^2 = \frac{1}{49}n^2$$

$$(b+3)(b-5)-(4+b)(b-4) = (b^2-2b-15)-(b^2-16)$$

$$= b^2-2b-15-b^2+16 = -2b+1$$