Question

помагитеееп пжжжжжж срочно​

Answer 1

Ответ:

$1) {(x + 6)}^{2} = {x}^{2} + 12x + 36 \\ 2) {(3a - 2)}^{2} = 9 {a}^{2} - 12a + 4 \\ 3)(3y - 2)(3y + 2) = 9 {y}^{2} - 4$

${(b - 8)}^{2} - {(64 - b}^{2} ) = {b}^{2} - 16b + 64 - 64 + {b}^{2} = 2 {b}^{2} - 16b = 2b(b - 8)$

$1)64 - {y}^{2} = (8 - y)(8 + y) \\ 2) {a}^{2} - 6ab + 9 {b}^{2} = {(a - 3)}^{2}$

$6 {x}^{2} - 24x = 0 \\ x(x - 24) = 0 \\ x1 = 0 \\ x2 = 24$

$1)36 {a}^{4} - 25 {a}^{2} {b}^{2} = {a}^{2} (36 {a}^{2} - 25 {b}^{2} ) = {a}^{2} (6a - 5b)(6a + 5b) \\ 2) {a}^{3} - 8 {b}^{3} = (a - 2b)( {a}^{2} + ab + 4{b}^{2} )$