Упр.15.15 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) (8a-3)(8a+3)-(7a+4)(8a-4);
2) 0,6m(2m-1)(2m+1)+0,3(6+5m)(6-5m);
3) (7-2x)(7+2x)-(x-8)(x+8)-(4-3x)(5+3x);
4)-b^2 c(4b-c^2 )(4b+c^2 )+16b^4 c.

1. $$\begin{aligned} (8a-3)(8a+3)-(7a+4)(8a-4) &= (64a^2-9)-(56a^2-28a+32a-16) \\ &= 64a^2-9-56a^2-4a+16 \\ &= 8a^2-4a+7. \end{aligned}$$

2. $$\begin{aligned} 0{,}6m(2m-1)(2m+1)+0{,}3(6+5m)(6-5m) &= 0{,}6m(4m^2-1)+0{,}3(36-25m^2) \\ &= 2{,}4m^3-0{,}6m+10{,}8-7{,}5m^2 \\ &= 2{,}4m^3-7{,}5m^2-0{,}6m+10{,}8. \end{aligned}$$

3. $$\begin{aligned} (7-2x)(7+2x)-(x-8)(x+8)-(4-3x)(5+3x) &= (49-4x^2)-(x^2-64)-(20+12x-15x-9x^2) \\ &= 49-4x^2-x^2+64-20+3x+9x^2 \\ &= 4x^2+3x+93. \end{aligned}$$

4. $$\begin{aligned} -b^2c(4b-c^2)(4b+c^2)+16b^4c &= -b^2c(16b^2-c^4)+16b^4c \\ &= -16b^4c+b^2c^5+16b^4c \\ &= b^2c^5. \end{aligned}$$

Ответ

1) $$8a^2-4a+7$$; 2) $$2{,}4m^3-7{,}5m^2-0{,}6m+10{,}8$$; 3) $$4x^2+3x+93$$; 4) $$b^2c^5$$.