Question

решите пожалуйста ❤
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Answer 1

1)

а) ${(x - 6)}^{2} = {x}^{2} - 12x + 36$

б) ${(3x + 1)}^{2} = 9 {x}^{2} + 6x + 1$

в) ${(2 {x}^{2} + {x}^{3}) }^{2} = {x}^{6} + 4 {x}^{5} + 4 {x}^{4}$

г) $(3y - 2)(3y + 2) = 9 {y}^{2} - 4$

д) $(4a + 3b)(3b - 4a) = 9 {b}^{2} - 16 {a}^{2}$

е) $( {c}^{2} - 3a)( {c}^{2} + 3a) = {c}^{4} - 9 {a}^{2}$

2) $(c - b)(c + b) - (5 {c}^{2} - {b}^{2} ) = {c}^{2} - {b}^{2} - 5 {c}^{2} + {b}^{2} = - 4 {c}^{2}$

3)

а) ${x}^{2} - 81 = (x - 9)(x + 9)$

б) ${x}^{2} + 4xy + 4 {y}^{2} = {(x + 2y)}^{2}$

в) $4 {x}^{2} {y}^{2} - 9 {a}^{4} = (2xy - 3 {a}^{2} )(2xy + 3 {a}^{2} )$

4)

$36 - {(6 - x)}^{2} = x(2.5 - x) \ 36 - (36 - 12x + {x}^{2} ) = 25x - {x}^{2} \ 36 - 36 + 12x - {x}^{2} = 25x - {x}^{2} \ - {x}^{2} + {x}^{2} - 25x + 12x = - 36 + 36 \ - 13x = 0 \ x = 0$

5) ${(3 + a)}^{2} {(3 - a)}^{2} = {((3 + a)(3 - a))}^{2} = {(9 - {a}^{2}) }^{2} = {a}^{4} - 18 {a}^{2} + 81$

6) ${( x - 7)}^{2} - 81 = (x - 16)(x +2)$