Упр.409 Часть 2 ГДЗ Дорофеев Петерсон 5 класс (Математика)

1) (1 4/7)^2 + 2 · 1 4/7 · 1 3/7 + (1 3/7)^2; 3) (32/141)^2 + 2 · 32/141 · 1 109/141 + (1 109/141)^2;
2) (4 9/11)^2 + 2 · 4 9/11 · 5 2/11 + (5 2/11)^2; 4) (3 5/12)^2 + 2 · 3 5/12 · 2 7/12 + (2 7/12)^2.

Используем формулу квадрата суммы:

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

1)

$$\left(1\frac{4}{7}\right)^2+2\cdot 1\frac{4}{7}\cdot 1\frac{3}{7}+\left(1\frac{3}{7}\right)^2=\left(1\frac{4}{7}+1\frac{3}{7}\right)^2$$

$$=\left(2+\frac{7}{7}\right)^2=3^2=9.$$

2)

$$\left(4\frac{9}{11}\right)^2+2\cdot 4\frac{9}{11}\cdot 5\frac{2}{11}+\left(5\frac{2}{11}\right)^2=\left(4\frac{9}{11}+5\frac{2}{11}\right)^2$$

$$=\left(9+\frac{11}{11}\right)^2=10^2=100.$$

3)

$$\left(\frac{32}{141}\right)^2+2\cdot \frac{32}{141}\cdot 1\frac{109}{141}+\left(1\frac{109}{141}\right)^2=\left(\frac{32}{141}+1\frac{109}{141}\right)^2$$

$$=\left(1+\frac{32+109}{141}\right)^2=\left(1+\frac{141}{141}\right)^2=2^2=4.$$

4)

$$\left(3\frac{5}{12}\right)^2+2\cdot 3\frac{5}{12}\cdot 2\frac{7}{12}+\left(2\frac{7}{12}\right)^2=\left(3\frac{5}{12}+2\frac{7}{12}\right)^2$$

$$=\left(5+\frac{12}{12}\right)^2=6^2=36.$$

Ответ

1) 9; 2) 100; 3) 4; 4) 36.