Question

Помогите пожалуйста выполнить домашнее задание номер 5.8 и 5.9 и другие номера

Answer 1

5.8

1)
$1 + {a}^{2} - 2a \: = {a}^{2} - 2a + 1 = {(a - 1)}^{2}$
2)
$4xy + {y}^{2} + {4x}^{2} = 4 {x}^{2} + 4xy + {y}^{2} = {(2x + y)}^{2}$
3)
$28ab + 49 {a}^{2} + 4 {b}^{2} = 49 {a}^{2} + 28ab + 4 {b}^{2} = {(7a + 2b)}^{2}$
4)
$10mn + 100 {m}^{2} + 0.25 {n}^{2} = 0.25 {n}^{2} + 10mn + 100 {m}^{2} = {(0.5n + 10m)}^{2}$
5)
$\frac{1}{4} {a}^{2} + 4 {b}^{2} - 2ab = \frac{1}{4} {a}^{2} - 2ab + 4 {b}^{2} = { (\frac{1}{2} a - 2b)}^{2}$
6)
$8ab + {b}^{2} + 16 {a}^{2} = 16 {a}^{2} + 8ab + {b}^{2} = {(4a + b)}^{2}$
5.9

1)
${a}^{2} + 2ac + {c}^{2} = {(a + c)}^{2}$
2)
${m}^{2} - 2mn + {n}^{2} = {(m - n)}^{2}$
3)
${x}^{2} - 4xy + 4{y}^{2} = {(x - 2y)}^{2}$
4)
${a}^{2} + 2ca + {c}^{2} = {(a + c)}^{2}$
5)
${c}^{2} + 14c + 49 = {(c + 7)}^{2}$
6)
${k}^{2} - 6k + 9 = {(k - 3)}^{2}$