Question

Формаула разности квадратов!
7. Разложите на множители:
а) 5x^2 - 125
б) 64 - 4y^2
в) ax^2 - ay^2
г) x^3 - xy^2
д) (x+y)^2 - z^2
е) 81 - (a + 2b)^2
ж) (a + 2b)^2 - (a - 2b)^2
з) (c + d)^2 - (2c + 3d)^2
и) a^2n - b^2m
к) c^4n - d^4

Answer 1

а)
$5 {x}^{2} - 125 = 5( {x}^{2} - 25) = 5(x - 5)(x + 5)$
б)
$64 - 4 {y}^{2} = 4(16 - {y}^{2} ) = 4(4 - y)(4 + y)$
в)
$a {x}^{2} - a {y}^{2} = a( {x}^{2} - {y}^{2} ) = a(x - y)(x + y)$
г)
${x}^{3} - x {y}^{2} = x( {x}^{2} - {y}^{2} ) = x(x - y)(x + y)$
д)
${(x + y)}^{2} - {z}^{2} = (x + y + z)(x + y - z)$
е)
$81 - {(a + 2b)}^{2} = (9 + a + 2b)(9 - a - 2b)$
ж)
${(a + 2b)}^{2} - {(a - 2b)}^{2} = (a + 2b + a - 2b)(a + 2b - a + 2b) = 2a \times 4b = 8ab$
з)
${(c + d)}^{2} - {(2c + 3d)}^{2} = (c + d + 2c + 3d)(c + d - 2c - 3d) = (3c + 4d)( - c - 2d)$
и)
${a}^{2n} - {b}^{2m} = ({a }^{n} - {b}^{m} )({a}^{n} + {b}^{n} )$
к)
${c}^{4n} - {d}^{4} = ( {c}^{2n} - {d}^{2} )( {c}^{2n} + {d}^{2} ) = ({c}^{n} - d)({c}^{n} + d)({c}^{2n} + d)$

Answer 2

а) 5x² - 125  = 5(x² - 25)  = 5(х-5)(х+5)

б) 64 - 4y² = 8² - (2у)²  = (8-2у)(8+2у)

в) ax² - ay²  = а(х² - у²) = а(х-у)(х+у)

г) x³ - xy² = х(х² - у²) = х(х-у)(х+у)

д) (x+y)² - z² = (х+у-z)(x+y+z)

е) 81 - (a + 2b)² =  9² - (a + 2b)² =  (9-a-2b)(9+a+2b)

ж) (a + 2b)² - (a - 2b)² = (a+2b-a+2b)(a+2b+a-2b)  = 4b · 2a = 8ab

з) (c + d)² - (2c + 3d)² =  (c+d-2c-3d)(c+d+2c+3d)  = (-c-2d)(3c+4d) =

 = -(c+2d)(3c+4d)

и)  $a^{2n} -b^{2m} = (a^n)^2-(b^m)^2=(a^n-b^m)(a^n+a^m)$

к) $c^{4n} - d^4 = (c^{2n})^2 - (d^2)^2 = (c^{2n}-d^2)(c^{2n}+d^2)=(c^{n}-d)(c^{n}+d)(c^{2n}+d^2)$