Question

24.58 (а,б) пж помогите

Answer 1

а)
$formula \: \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ (3 {c}^{2} + 4 {d}^{4} )^{2} = 9 {c}^{4} + 24 {c}^{2} {d}^{4} + 16 {d}^{8}$
▪Решение
рассуждаем так:
1) в формуле
$2xy = 24 {c}^{2} {d}^{4} \\ y = 4d ^{4}$
2) значит:
$2x \times 4 {d}^{4} = 24 {c}^{2} {d}^{4} \: \: \: \: | \div 8 {d}^{4} \\ x = 3 {c}^{2}$
б) Решение:
$formula \: \: \: \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ 1) \: \: \: {x}^{2} = 81 {a}^{6} {b}^{2} \\ x = \sqrt{81 {a}^{6} {b}^{2} } \\ x = 9 {a}^{3} b \\ - - - - - - - \\ 2) \: \: \: y = 8 {a}^{4} \\ - - - - - - - \\ 3) \: \: \: {(9 {a}^{3} b - 8 {a}^{4} )}^{2} = 81 {a}^{6} {b}^{2} - 144 {a}^{7} b + 64 {a}^{8}$