Question

помогите плз 100 баллов даи​

Answer 1

1) 36 - 16y² = 6² - (4y)² = (6 - 4y)(6 + 4y) = 2 * 2 * (3 - 2y)(3 + 2y) =

= 4(3 - 2y)(3 + 2y)

2) 4x² - 81y² = (2x)² - (9y)² = (2x - 9y)(2x + 9y)

3) 0,04t² - 169p² = (0,2t)² - (13p)² = (0,2t - 13p)(0,2t + 13p)

4) x⁶ - x¹⁰ = x⁶(1 - x⁴) = x⁶(1 - x²)(1 + x²)

5) 25a⁶b² - 1 = (5a³b)² - 1² = (5a³b - 1)(5a³b + 1)

Answer 2

$36 - 16y {}^{2} = 4(9 - 4y {}^{2} ) = 4(3 - 2y)(3 + 2y)$

$4x{}^{2} - 81y {}^{2} = 2 {}^{2} {x}^{2} - {9}^{2} {y}^{2} = (2x {)}^{2} - (9y {)}^{2} = (2x - 9y)(2x + 9y)$

$0.04 {t}^{2} - 169 {p}^{2} = \frac{1}{25} {t}^{2} - 169 {p}^{2} = ( \frac{1}{5} t - 13p)( \frac{1}{5} t + 13p) = (0.2t - 13p)(0.2t + 13p)$

${x}^{6} - {x}^{10} = {x}^{6} (1 - x {}^{4} ) = {x}^{6} (1 - x {}^{2} )( 1+ {x}^{2} ) = {x}^{6} (1 - x)(1 + x)(1 + {x}^{2} )$

$25a {}^{6} b {}^{2} - 1 = {5}^{2} {a}^{2 \times 3} b {}^{2} - {1}^{2} = {5}^{2} (a {}^{3} {)}^{2} \times b {}^{2} - {1}^{2} = (5 {a}^{3} {b)}^{2} - {1}^{2} = (5a {}^{3} b - 1)(5a {}^{3} b + 1)$