Упр.15.4 ГДЗ Мерзляк Поляков 7 класс (Алгебра)
- 15.4. Выполните умножение:
1) $$\left(a^2-3\right)\left(a^2+3\right)$$;
2) $$\left(5+b^2\right)\left(b^2-5\right)$$;
3) $$\left(3x-2y^2\right)\left(3x+2y^2\right)$$;
4) $$\left(10p^3-7k\right)\left(10p^3+7k\right)$$;
5) $$\left(4x^2-8y^3\right)\left(4x^2+8y^3\right)$$;
6) $$\left(11a^3+5b^2\right)\left(5b^2-11a^3\right)$$;
7) $$\left(7-xy\right)\left(7+xy\right)$$;
8) $$\left(8a^3b-\frac{1}{3}ab^2\right)\left(8a^3b+\frac{1}{3}ab^2\right)$$;
9) $$\left(0{,}3m^5+0{,}1n^3\right)\left(0{,}3m^5-0{,}1n^3\right)$$;
10) $$\left(\frac{7}{9}a^2c-1{,}4b^4\right)\left(1{,}4b^4+\frac{7}{9}a^2c\right)$$.
Используем формулу разности квадратов:
$$(u-v)(u+v)=u^2-v^2.$$
$$(a^2-3)(a^2+3)=(a^2)^2-3^2=a^4-9$$
$$(5+b^2)(b^2-5)=(b^2+5)(b^2-5)=(b^2)^2-5^2=b^4-25$$
$$(3x-2y^2)(3x+2y^2)=(3x)^2-(2y^2)^2=9x^2-4y^4$$
$$(10p^3-7k)(10p^3+7k)=(10p^3)^2-(7k)^2=100p^6-49k^2$$
$$(4x^2-8y^3)(4x^2+8y^3)=(4x^2)^2-(8y^3)^2=16x^4-64y^6$$
$$(11a^3+5b^2)(5b^2-11a^3)=(5b^2+11a^3)(5b^2-11a^3)$$
$$=(5b^2)^2-(11a^3)^2=25b^4-121a^6$$
$$(7-xy)(7+xy)=7^2-(xy)^2=49-x^2y^2$$
$$\left(8a^3b-\frac13ab^2\right)\left(8a^3b+\frac13ab^2\right) =\left(8a^3b\right)^2-\left(\frac13ab^2\right)^2$$
$$=64a^6b^2-\frac19a^2b^4$$
$$(0{,}3m^5+0{,}1n^3)(0{,}3m^5-0{,}1n^3) =(0{,}3m^5)^2-(0{,}1n^3)^2$$
$$=0{,}09m^{10}-0{,}01n^6$$
$$\left(\frac79a^2c-1{,}4b^4\right)\left(1{,}4b^4+\frac79a^2c\right) =\left(\frac79a^2c-1{,}4b^4\right)\left(\frac79a^2c+1{,}4b^4\right)$$
$$=\left(\frac79a^2c\right)^2-(1{,}4b^4)^2 =\frac{49}{81}a^4c^2-1{,}96b^8$$
Ответ
1) $$a^4-9$$; 2) $$b^4-25$$; 3) $$9x^2-4y^4$$; 4) $$100p^6-49k^2$$; 5) $$16x^4-64y^6$$; 6) $$25b^4-121a^6$$; 7) $$49-x^2y^2$$; 8) $$64a^6b^2-\frac19a^2b^4$$; 9) $$0{,}09m^{10}-0{,}01n^6$$; 10) $$\frac{49}{81}a^4c^2-1{,}96b^8$$.
