初一奥数讲座因式分解(1)答案

初一奥数讲座 因式分解（1） 例1．分解因式（提公因式法） （1）4a2 + 6ab + 2a 解：原式 = 2a(2a + 3b + 1) （2）2am + 1 + 4am – 2am – 1 解：原式 = 2am – 1(a2 + 2a – 1) （3）(m – n) – (n – m)2 解：原式 = (m – n)2 – (m – n)2 = (m – n)[1 – (m – n)] = (m – n)(1 – m + n) （4）2a2b(b + c)(x + y)2 – 6a3b2(b + c)2(x + y) 解：原式 = 2a2b(b + c)(x + y)[(x + y) – 3ab(b + c)] = 2a2b(b + c)(x + y)(x + y – 3ab2 – 3abc) 例2．分解因式（运用公式法） （1）x2 – 81 解：原式 = x2 – 92 = (x + 9)(x – 9) （2）4(x + y)2 – 9(x – y)2 解：原式 = [2(x + y) + 3(x – y)][2(x + y) – 3(x – y)] = (5x – y)(– x + 5y) = – (5x – y)(x – 5y) （3）x2 + 8xy + 16y2 解：原式 = x2 + 2·x·4y + (4y)2 = (x + 4y)2 （4）(x2 – 2x)2 + 2(x2 – 2x) + 1 解：原式 = (x2 – 2x)2 + 2(x2 – 2x)·1 + 12 = (x2 – 2x + 1)2 = [(x – 1)2]2 = (x – 1)4 例3．分解因式（运用公式法） （1）125a3b6 + 8 解：原式 = (5ab2)3 + 23 = (5ab2 + 2)[(5ab2)2 – 2×5ab2 + 22] = (5ab2 + 2)(25a2b4 – 10ab2 + 4) （2）512x9 – 1 解：原式 = (8x3)3 – 13 = (8x3 – 1)[(8x3)2 + 8x3 + 1] = (2x – 1)(4x2 + 2x + 1)(64x6 + 8x3 + 1) （3）1 – 12x2y2 + 48x4y4 – 64x6y6 解：原式 = 1 – 3×4x2y2 + 3×(4x2y2)2 – (4x2y2)3 = (1 – 4x2y2)3 = (1 + 2xy)3(1 – 2xy)3 （4）x3 + 3xy + y3 – 1 解：原式 = x3 + y3 + (– 1)3 – 3·x·y(– 1) = (x + y – 1)(x2 + y2 + 1 – xy + y + x) （5）x2 + 9y2 + 4z2 – 6xy + 4xz – 12yz 解：原式 = x2 + (– 3y)2 + (– 2z)2 + 2·x·(– 3y) + 2·x·2z + 2·(– 3y)·(2z) = (x – 3y + 2z)2 例4．分解因式 （1） x2 – xy + y2 解：原式 = (x2 – 2xy + y2) = (x – y)2 （2）100 – 25x2 解：原式 = 25(4 – x2) = 25(2 + x)(2 – x) （3）x4 – 2x2y2 + y4 解：原式 = (x2)2 – 2x2y2 + (y2)2 = (x2 – y2)2 = (x + y)2(x – y)2 （4）2a6 – a3 + 解：原式 = 2(a6 – a3 + ) = 2[(a3)2 – 2× ·a3 + ( )2] = 2(a3 – )2 = 2(a – )2(a2 + a + ) 例5．分解因式 （1）– 2x5n – 1yn + 4x3n – 1yn + 2 – 2xn – 1yn + 4 解：原式 = – 2xn – 1yn(x4n – 2x2ny2 + y4) = – 2xn – 1yn[(x2n)2 – 2x2ny2 + (y2)2] = – 2xn – 1yn(x2n – y2)2 = – 2xn – 1yn(xn + y)(xn – y) （2）(a2 + ab + b2)2 – 4ab(a2 + b2) 解：原式 = [(a2 + b2) + ab]2 – 4ab(a2 + b2) = (a2 + b2)2 + 2ab(a2 + b2) + a2b2 – 4ab(a2 + b2) = (a2 + b2)2 – 2ab(a2 + b2) + a2b2 = (a2b2 – ab)2 （3）(x2 – x) – 4(x – 2)(x + 1) – 4 解：原式 = (x2 – x)2 – 4(x2 – x – 2) – 4 = (x2 – x)2 – 4(x2 – x) + 8 – 4 = (x2 – x)2 – 4(x2 – x) + 4 = (x2 – x – 2)2 = (x – 2)2(x + 1)2 （4）a7 – a5b2 + a2b5 – b7 解：原式 = (a7 – a5b2) + (a2b5 – b7) = a5(a2 – b2) + b5(a2 – b2) = (a2 – b2)(a5 + b5) = (a + b)(a – b)(a + b)(a4 – a3b + a2b2 – ab3 + b4) = (a + b)2(a – b)(a4 – a3b + a2b2 – ab3 + b4) 例6．分解因式 （1）a3 + b3 + c3 – 3abc 解：原式 = (a + b)3 – 3ab(a + b) + c3 – 3abc = [(a + b)3 + c3] – 3ab(a + b + c) = (a + b + c)[(a + b)2 – (a + b)c + c2] – 3ab(a + b + c) = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) （2）(x + y)3 + (z – x)3 – (y + z)3 解：原式 = [(x + y) + (z – x)][(x + y)2 – (x + y)(z – x) + (z – x)2] – (y + z)3 = (y + z)[(x + y)2 – (x + y)(z – x) + (z – x)2 –(y + z)2] = (y + z)(3x2 + 3xy – 3yz – 3xz) = 3(y + z)[x(x + y) – z(x + y)] = 3(y + z)(x + y)(x – z) （3）x15 + x14 + x13 + … + x2 + x + 1 解：因为x16 – 1 = (x15 + x14 + x13 + … + x2 + x + 1) ∴原式 = = = = (x8 + 1)(x4 + 1)(x2 + 1)(x + 1) 例7．分解因式（分组分解法） （1）a2 – b2 – 2a – 2b 解：原式 = (a + b)(a – b) – 2(a + b) = (a + b)(a – b – 2) （2）25a4 – x2 – 2x – 1 解：原式 = (5a2)2 – (x2 + 2x + 1) = (5a2)2 – (x + 1)2 = (5a2 + x + 1)(5a2 – x – 1) （3）4a2 – b2 – 2a + 解：原式 = 4a2 – 2a + – b2 = (2a – )2 – b2 = (2a – + b)( 2a – – b) （4）(1 – a2)(1 – b2) – 4ab 解：原式 = 1 – a2 – b2 + a2b2 – 4ab = a2b2 – 2ab + 1 – a2 – 2ab – b2 = (ab – 1)2 – (a + b)2 = (ab – 1 + a + b)(ab – 1 – a – b) （5）a4 + a2b2 + b4 解：原式 = a4 + 2a2b2 + b4 – a2b2 = (a2 + b2)2 – a2b2 = (a2 + b2 + ab)( a2 + b2 – ab) 练习 1．：817 – 279 – 913能被45整除 证明：∵817 – 279 – 913 = 328 – 327 – 326 = 326(32 – 3 – 1) = 326×5 = 324×32×5 = 324×45 ∴817 – 279 – 913能被45整除 2．求证：四个连续自然数的积再加上1，一定是一个完全平方数 证明：设这四个连续自然数分别为n，n + 1，n + 2，n + 3 n(n + 1)(n + 2)(n + 3) + 1 = n(n + 3) (n + 1)(n + 2) + 1 = (n2 + 3n)(n2 + 3n + 1) + 1 = (n2 + 3n)2 + 2(n2 + 3n) + 1 = (n2 + 3n + 1)2 ∴n(n + 1)(n + 2)(n + 3) + 1一定是一个完全平方数 PAGE 4 _1297486429.unknown _1297486698.unknown _1297488771.unknown _1297489066.unknown _1297489093.unknown _1297488780.unknown _1297488696.unknown _1297486627.unknown _1297486670.unknown _1297486461.unknown _1297486599.unknown _1297236301.unknown _1297236362.unknown _1297238910.unknown _1297236352.unknown _1297236285.unknown