Упр.354 ГДЗ Никольский Потапов 7 класс (Алгебра)

а) (1/5*m-m3)2;
б) (-1/2+3bc)2;
в) (1/2*x3-1/3*y4)2;
г) (-1*1/2*p2+2/3*q)2;
д) (1*1/3*ab2-3a2b)2;
е) (2m3n2-1*1/2*mn3)2;
ж) (0,1а + 3а2b)2;
з) (2m3n2-2*1/2*mn3)2;
и) (-0,5х3у2 + 0,3ху5)2.

Используем формулу квадрата суммы и квадрата разности:

$$ (a \pm b)^2 = a^2 \pm 2ab + b^2. $$

1. $$\left(\frac{1}{5}mn-m^3\right)^2=\left(\frac{1}{5}mn\right)^2-2\cdot \frac{1}{5}mn\cdot m^3+(m^3)^2$$

   $$=\frac{1}{25}m^2n^2-\frac{2}{5}m^4n+m^6.$$

2. $$\left(-\frac{1}{2}+3bc\right)^2=\left(-\frac{1}{2}\right)^2-2\cdot \frac{1}{2}\cdot 3bc+(3bc)^2$$

   $$=\frac{1}{4}-3bc+9b^2c^2.$$

3. $$\left(\frac{1}{2}x^3-\frac{1}{3}y^4\right)^2=\left(\frac{1}{2}x^3\right)^2-2\cdot \frac{1}{2}x^3\cdot \frac{1}{3}y^4+\left(\frac{1}{3}y^4\right)^2$$

   $$=\frac{1}{4}x^6-\frac{1}{3}x^3y^4+\frac{1}{9}y^8.$$

4. $$\left(-1\frac{1}{2}p^2+\frac{2}{3}q\right)^2=\left(-\frac{3}{2}p^2\right)^2-2\cdot \frac{3}{2}p^2\cdot \frac{2}{3}q+\left(\frac{2}{3}q\right)^2$$

   $$=\frac{9}{4}p^4-2p^2q+\frac{4}{9}q^2.$$

5. $$\left(1\frac{1}{3}ab^2-3a^2b\right)^2=\left(\frac{4}{3}ab^2\right)^2-2\cdot \frac{4}{3}ab^2\cdot 3a^2b+(3a^2b)^2$$

   $$=\frac{16}{9}a^2b^4-8a^3b^3+9a^4b^2.$$

6. $$\left(2m^3n^2-1\frac{1}{2}mn^3\right)^2=\left(2m^3n^2\right)^2-2\cdot 2m^3n^2\cdot \frac{3}{2}mn^3+\left(\frac{3}{2}mn^3\right)^2$$

   $$=4m^6n^4-6m^4n^5+\frac{9}{4}m^2n^6.$$

7. $$\left(0{,}1a+3a^2b\right)^2=(0{,}1a)^2+2\cdot 0{,}1a\cdot 3a^2b+(3a^2b)^2$$

   $$=0{,}01a^2+0{,}6a^3b+9a^4b^2.$$

8. $$\left(2m^3n^2-2\frac{1}{2}mn^3\right)^2=\left(2m^3n^2\right)^2-2\cdot 2m^3n^2\cdot \frac{5}{2}mn^3+\left(\frac{5}{2}mn^3\right)^2$$

   $$=4m^6n^4-10m^4n^5+\frac{25}{4}m^2n^6.$$

9. $$\left(-0{,}5x^3y^2+0{,}3xy^5\right)^2=\left(-0{,}5x^3y^2\right)^2-2\cdot 0{,}5x^3y^2\cdot 0{,}3xy^5+\left(0{,}3xy^5\right)^2$$

   $$=0{,}25x^6y^4-0{,}3x^4y^7+0{,}09x^2y^{10}.$$

Ответ

а) $$\frac{1}{25}m^2n^2-\frac{2}{5}m^4n+m^6$$;
б) $$\frac{1}{4}-3bc+9b^2c^2$$;
в) $$\frac{1}{4}x^6-\frac{1}{3}x^3y^4+\frac{1}{9}y^8$$;
г) $$\frac{9}{4}p^4-2p^2q+\frac{4}{9}q^2$$;
д) $$\frac{16}{9}a^2b^4-8a^3b^3+9a^4b^2$$;
е) $$4m^6n^4-6m^4n^5+\frac{9}{4}m^2n^6$$;
ж) $$0{,}01a^2+0{,}6a^3b+9a^4b^2$$;
з) $$4m^6n^4-10m^4n^5+\frac{25}{4}m^2n^6$$;
и) $$0{,}25x^6y^4-0{,}3x^4y^7+0{,}09x^2y^{10}$$.