Homepage › Solution manuals › Найма Гахраманова › Математика - 8 › Задание 9, стр 101
a) b−a a ⋅ 3a (a2−b2)b) (a2−1) a−b ⋅3a−3b) a2+a) .c) 6a x2−x : 3ax 2x−2d) (a+2)2 (a2−9) ⋅2a−6 2a+4.e) c2+2c c2−1 : 2c+4 3c−3f) (x+1)2 2x+4 : (x2−1) 3x+6 .
a) b−a a ⋅ 3a (a2−b2) = b−a a ⋅ 3a (a−b)(a+b) = −a−b a ⋅ 3a (a−b)(a+b) = − 3 a+b.b) (a2−1) a−b ⋅3a−3b) a2+a) = (a−1)(a+1) a−b ⋅3(a−b) a(a+1) = 3(a−1) a = 3a−3 a .c) 6a x2−x : 3ax 2x−2 = 6a x(x−1) ⋅2x−2 3ax = 3a⋅2 x(x−1) ⋅2(x−1) 3a⋅x = 2⋅2 x⋅x = 4 x2;d) (a+2)2 (a2−9) ⋅2a−6 2a+4 = (a+2)(a+2) (a−3)(a+3) ⋅2(a−3) 2(a+2) = a+2 a+3.e) c2+2c c2−1 : 2c+4 3c−3 = c2+2c c2−1 ⋅3c−3 2c+4 = c(c+2) (c−1)(c+1) ⋅3(c−1) 2(c+2) = 3c 2(c+1).f) (x+1)2 2x+4 : (x2−1) 3x+6 = (x+1)2 2(x+2) ⋅3x+6 x2−1 = (x+1)(x+1) 2(x+2) ⋅ 3(x+2) (x−1)(x+1) = (x+1)⋅3 2(x−1) = 3x+3 2x−2.