Упр.875 ГДЗ Макарычев 7 класс (Алгебра)
- Задание учебника 2026 года.
Представьте в виде многочлена:
а) $$\left(3x^2-1\right)\left(3x^2+1\right)$$;
б) $$\left(5a-b^3\right)\left(b^3+5a\right)$$;
в) $$\left(\frac{3m^3}{7}+\frac{n^3}{4}\right)\left(\frac{3m^3}{7}-\frac{n^3}{4}\right)$$;
г) $$\left(\frac{1}{15}-\frac{p^6}{8}\right)\left(\frac{p^6}{8}+\frac{1}{15}\right)$$;
д) $$\left(0{,}4p^3+5a^2\right)\left(5a^2-0{,}4y^3\right)$$;
е) $$\left(1{,}2c^2-7a^2\right)\left(1{,}2c^2+7a^2\right)$$;
ж) $$\left(\frac{5x}{8}+y^5\right)\left(y^5-\frac{5x}{8}\right)$$;
з) $$\left(\frac{2p^5}{7}-0{,}01\right)\left(0{,}01+\frac{p^5}{7}\right)$$. - Задание учебника 2019 года.
Упростите выражение:
а) $$5a(a-8)-3(a+2)(a-2)$$;
б) $$(1-4b)(4b+1)+6b(b-2)$$;
в) $$(8p-q)(q+8p)-(p+q)(p-q)$$;
г) $$(2x-7y)(2x+7y)+(2x-7y)(7y-2x)$$.
Решение
$$\left(3x^2-1\right)\left(3x^2+1\right)=\left(3x^2\right)^2-1^2=9x^4-1$$
$$\left(5a-b^3\right)\left(b^3+5a\right)=\left(5a-b^3\right)\left(5a+b^3\right)=\left(5a\right)^2-\left(b^3\right)^2=25a^2-b^6$$
$$\left(\frac{3}{7}m^3+\frac{1}{4}n^3\right)\left(\frac{3}{7}m^3-\frac{1}{4}n^3\right)=\left(\frac{3}{7}m^3\right)^2-\left(\frac{1}{4}n^3\right)^2=\frac{9}{49}m^6-\frac{1}{16}n^6$$
$$\left(\frac{1}{15}-\frac{1}{8}p^6\right)\left(\frac{1}{8}p^6+\frac{1}{15}\right)=\left(\frac{1}{15}-\frac{1}{8}p^6\right)\left(\frac{1}{15}+\frac{1}{8}p^6\right)=\left(\frac{1}{15}\right)^2-\left(\frac{1}{8}p^6\right)^2=\frac{1}{225}-\frac{1}{64}p^{12}$$
$$\left(0{,}4y^3+5a^2\right)\left(5a^2-0{,}4y^3\right)=\left(5a^2+0{,}4y^3\right)\left(5a^2-0{,}4y^3\right)=\left(5a^2\right)^2-\left(0{,}4y^3\right)^2$$
$$=25a^4-0{,}16y^6$$
$$\left(1{,}2c^2-7a^2\right)\left(1{,}2c^2+7a^2\right)=\left(1{,}2c^2\right)^2-\left(7a^2\right)^2=1{,}44c^4-49a^4$$
$$\left(\frac{5}{8}x+y^5\right)\left(y^5-\frac{5}{8}x\right)=\left(y^5+\frac{5}{8}x\right)\left(y^5-\frac{5}{8}x\right)=\left(y^5\right)^2-\left(\frac{5}{8}x\right)^2$$
$$=y^{10}-\frac{25}{64}x^2$$
$$\left(\frac{2}{7}p^5-0{,}01\right)\left(0{,}01+\frac{1}{7}p^5\right)=\left(\frac{1}{7}p^5-0{,}01\right)\left(0{,}01+\frac{1}{7}p^5\right)$$
$$=\left(\frac{1}{7}p^5\right)^2-\left(0{,}01\right)^2=\frac{1}{49}p^{10}-0{,}0001$$
Ответ
а) $$9x^4-1$$; б) $$25a^2-b^6$$; в) $$\frac{9}{49}m^6-\frac{1}{16}n^6$$; г) $$\frac{1}{225}-\frac{1}{64}p^{12}$$; д) $$25a^4-0{,}16y^6$$; е) $$1{,}44c^4-49a^4$$; ж) $$y^{10}-\frac{25}{64}x^2$$; з) $$\frac{1}{49}p^{10}-0{,}0001$$.
