Question

Решите пжжжжжжжжжжжжжжж

Answer 1

$4 - 4x + {x}^{2} = (2 - x {)}^{2} \\64 + 16b + {b}^{2} = (8 + b {)}^{2} \\ {x}^{2} - 14x + 49 = (x - 7 {)}^{2} \\ {a}^{2} + 81 - 18a = {a}^{2} - 18a + 81 = (a - 9 {)}^{2} \\2cd + {c}^{2} + {d}^{2} = {c}^{2} + 2cd + {d}^{2} = (c + d {)}^{2}$

$81 {a}^{2} - 18ab + {b}^{2} = (9a - b {)}^{2} \\ {b}^{2} + 4 {a}^{2} - 4ab =(b - 2a {)}^{2} \\ 4 {x}^{4} - 12 {x}^{2} + 9 = (2 {x}^{2} - 3{)}^{2} \\ 9 + 6 {a}^{2} b + {a}^{4} {b}^{2} = (3 + {a}^{2}b {)}^{2} \\ 4 {y}^{2} - 20yz + 25 {z}^{2} = (2y - 5z {)}^{2}$

$24 {x}^{2} {y}^{2} - 9 {x}^{4} - 16 {y}^{4} = - (3 {x}^{2} - 4 {y}^{2} {)}^{2} \\ {m}^{4} - 20 {m}^{2} n + 100 {n}^{2} =( {m}^{2} - 10n {)}^{2} \\ 0.09 - 3 {b}^{3} + 25 {b}^{6} = (0.3 - 5 {b}^{3} {)}^{2} \\ 0.49 {c}^{4} + 1.4 {c}^{2} +1 = (0.7 {c}^{2} + 1 {)}^{2} .$