Question

ПОМОГИТЕ ПОЖАЛУЙСТА РЕШИТЬ АЛГЕБРУ 7 КЛАСС

Answer 1

1) $27 - x^3 = 3^3 - x^3 = (3 - x)\cdot (3^2 + 3x + x^2) = (3-x)\cdot (9 + 3x + x^2)$

2) $a^3 + 64 = a^3 + 4^3 = (a + 4)\cdot (a^2 - 4a + 4^2) = (a+4)\cdot (a^2 - 4a + 16)$

3) $8x^3 - y^3 = (2x)^3 - y^3 = (2x - y)\cdot ( (2x)^2 + 2xy + y^2) =$

$= (2x - y)\cdot (4x^2 + 2xy + y^2)$

4) $216 - m^3 n^3 = 6^3 - (mn)^3 = (6 - mn)\cdot ( 6^2 + 6mn + (mn)^2 ) =$

$= (6 - mn)\cdot ( 36 + 6mn + m^2 n^2)$

5) $b^9 + a^{12} = (b^3)^3 + (a^4)^3 = (b^3 + a^4)\cdot ((b^3)^2 - b^3 a^4 + (a^4)^2) =$

$= (b^3 + a^4)\cdot ( b^6 - b^3 a^4 + a^8 )$

6) $343 a^6 b^{15} - 0{,}008 x^9 y^3 = (7a^2b^5)^3 - (0{,}2x^3y)^3 =$

$= ( 7a^2b^5 - 0{,}2x^3y )\cdot ( (7a^2b^5)^2 + (7a^2b^5)\cdot(0{,}2x^3y) +$

$+ (0{,}2x^3y)^2 ) = ( 7a^2b^5 - 0{,}2x^3y )\cdot ( 49a^4b^{10} + 1{,}4a^2b^5x^3y +$

$+ 0{,}04x^6y^2 )$