Упр.15.4 ГДЗ Мерзляк Поляков 7 класс (Алгебра)
1) (a^2-3)(a^2+3);
2) (5+b^2 )(b^2-5);
3) (3x-2y^2 )(3x+2y^2 );
4) (10p^3-7k)(10p^3+7k);
5) (4x^2-8y^3 )(4x^2+8y^3 );
6) (11a^3+5b^2 )(5b^2-11a^3 );
7) (7-xy)(7+xy);
8) (8a^3 b-1/3 ab^2 )(8a^3 b+1/3 ab^2 );
9) (0,3m^5+0,1n^3 )(0,3m^5-0,1n^3 );
10) (7/9 a^2 c-1,4b^4 )(1,4b^4+7/9 a^2 c).
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) $$\frac{49}{81}a^4c^2-1{,}96b^8$$.
