分解因式
(1). p2(a−1)+p(1−a)p^2(a-1)+p(1-a)p2(a−1)+p(1−a)
= p2(a−1)−(a−1)p^2(a-1)-(a-1)p2(a−1)−(a−1)
= p(a−1)(p−1)p(a-1)(p-1)p(a−1)(p−1)
(2). (a2−a)2−14(a2−a)+24(a^2-a)^2-14(a^2-a)+24(a2−a)2−14(a2−a)+24
= (a2−a−2)(a2−a−12)(a^2-a-2)(a^2-a-12)(a2−a−2)(a2−a−12)
= (a−2)(a+1)(a−4)(a+3)(a-2)(a+1)(a-4)(a+3)(a−2)(a+1)(a−4)(a+3)
(3). ab+b2+a−b−2ab+b^2+a-b-2ab+b2+a−b−2
= b(a+b)+(a+b)−2(b+1)b(a+b)+(a+b)-2(b+1)b(a+b)+(a+b)−2(b+1)
= (b+1)(a+b)−2(b+1)(b+1)(a+b)-2(b+1)(b+1)(a+b)−2(b+1)
= (b+1)(a+b−2)(b+1)(a+b-2)(b+1)(a+b−2)
(4). 2x4+13x3+20x2+11x+22x^4+13x^3+20x^2+11x+22x4+13x3+20x2+11x+2
= 2x4+13x3+6x2+14x2+11x+22x^4+13x^3+6x^2+14x^2+11x+22x4+13x3+6x2+14x2+11x+2
= x2(2x+1)(x+6)+(2x+1)(7x+2)x^2(2x+1)(x+6)+(2x+1)(7x+2)x2(2x+1)(x+6)+(2x+1)(7x+2)
= (2x+1)(x3+6x2+7x+2)(2x+1)(x^3+6x^2+7x+2)(2x+1)(x3+6x2+7x+2)
= (2x+1)[x(x2+6x+5)+2(x+1)](2x+1)[x(x^2+6x+5)+2(x+1)](2x+1)[x(x2+6x+5)+2(x+1)]
= (2x+1)(x+1)(x2+5x+2)(2x+1)(x+1)(x^2+5x+2)(2x+1)(x+1)(x2+5x+2)
(5). 6x2+xy−15y2+4x−25y−106x^2+xy-15y^2+4x-25y-106x2+xy−15y2+4x−25y−10
= (2x−3y)(3x+5y)+(4x−25y)−10(2x-3y)(3x+5y)+(4x-25y)-10(2x−3y)(3x+5y)+(4x−25y)−10
= (2x−3y−2)(3x+5y+5)(2x-3y-2)(3x+5y+5)(2x−3y−2)(3x+5y+5)
(6). 如果把多项式 x2−8x+mx^2-8x+mx2−8x+m 分解因式得 (x−10)(x+n)(x-10)(x+n)(x−10)(x+n),求m、nm、nm、n的值。
(x−10)(x+n)=x2+(10−n)x−10n(x-10)(x+n)=x^2+(10-n)x-10n(x−10)(x+n)=x2+(10−n)x−10n
n−10=−8n-10=-8n−10=−8
−10n=m-10n=m−10n=m
n=−2n=-2n=−2
m=20m=20m=20
(7). 已知 3a+3b=−93a+3b=-93a+3b=−9,求 2a2+4ab+2b2−62a^2+4ab+2b^2-62a2+4ab+2b2−6 的值。
a+b=−3a+b=-3a+b=−3
2a2+4ab+2b2−62a^2+4ab+2b^2-62a2+4ab+2b2−6
= 2(a+b)2−62(a+b)^2-62(a+b)2−6
= 121212
(8). 设 a(a−1)−(a2−b)=2a(a-1)-(a^2-b)=2a(a−1)−(a2−b)=2 ,求 a2+b22−ab\frac {a^2+b^2}2-ab2a2+b2−ab 的值。
a(a−1)−(a2−b)=2a(a-1)-(a^2-b)=2a(a−1)−(a2−b)=2
= a2−a−a2+ba^2-a-a^2+ba2−a−a2+b
= −a+b=2-a+b=2−a+b=2
a2+b22−ab=a2+b2−2ab2=(a−b)22=2\frac {a^2+b^2}2-ab=\frac {a^2+b^2-2ab}2=\frac{(a-b)^2}{2}=22a2+b2−ab=2a2+b2−2ab=2(a−b)2=2
1.把下列各式分解因式:
(1) x2+2y−y2−2xx^2+2y-y^2-2xx2+2y−y2−2x
= (x2−y2)−(2x−2y)(x^2-y^2)-(2x-2y)(x2−y2)−(2x−2y)
= (x+y)(x−y)−2(x−y)(x+y)(x-y)-2(x-y)(x+y)(x−y)−2(x−y)
= (x+y−2)(x−y)(x+y-2)(x-y)(x+y−2)(x−y)
(2) 2a2+bc2−a2b−2ac2a^2+bc^2-a^2b-2ac2a2+bc2−a2b−2ac
= (2a2−2ac)+(bc2−a2b)(2a^2-2ac)+(bc^2-a^2b)(2a2−2ac)+(bc2−a2b)
= 2a(a−c)−b(a2−c2)2a(a-c)-b(a^2-c^2)2a(a−c)−b(a2−c2)
= 2a(a−c)−b(a−c)(a+c)2a(a-c)-b(a-c)(a+c)2a(a−c)−b(a−c)(a+c)
= (a−c)(2a−ab−bc)(a-c)(2a-ab-bc)(a−c)(2a−ab−bc)
(3) a2+b2+2ac+2bc+2aba^2+b^2+2ac+2bc+2aba2+b2+2ac+2bc+2ab
= (a2+2ab+b2)+(2ac+2bc)(a^2+2ab+b^2)+(2ac+2bc)(a2+2ab+b2)+(2ac+2bc)
= (a+b)2+2c(a+b)(a+b)^2+2c(a+b)(a+b)2+2c(a+b)
= (a+b)(a+b+2c)(a+b)(a+b+2c)(a+b)(a+b+2c)
(4) 4x2+y2−z2−4xy4x^2+y^2-z^2-4xy4x2+y2−z2−4xy
= (2x−y)2−z2(2x-y)^2-z^2(2x−y)2−z2
= (2x−y+z)(2x−y−z)(2x-y+z)(2x-y-z)(2x−y+z)(2x−y−z)
2.把下列关于 xxx 的二次多项式因式分解
(1) x2+x−1x^2+x-1x2+x−1
= (x+1+52)(x−1+52)(x+\frac {1+\sqrt 5}2)(x-\frac {1+\sqrt 5}2)(x+21+5)(x−21+5)
(2) x2+4xy−4y2x^2+4xy-4y^2x2+4xy−4y2
= [x+2(1−2)y][x+2(1+2)y][x+2(1-\sqrt 2)y][x+2(1+\sqrt 2)y][x+2(1−2)y][x+2(1+2)y]
1.分解因式: x2−y2−2y−1x^2-y^2-2y-1x2−y2−2y−1
= x2−(y2+2y+1)x^2-(y^2+2y+1)x2−(y2+2y+1)
= x2−(y+1)2x^2-(y+1)^2x2−(y+1)2
= (x+y+1)(x−y−1)(x+y+1)(x-y-1)(x+y+1)(x−y−1)
2.把 2(a2−3mn)+a(4m−3n)2(a^2-3mn)+a(4m-3n)2(a2−3mn)+a(4m−3n) 因式分解
= 2a2−6mn+4am−3an2a^2-6mn+4am-3an2a2−6mn+4am−3an
= 2a(a+2m)−3n(a+2m)2a(a+2m)-3n(a+2m)2a(a+2m)−3n(a+2m)
= (a+2m)(2a−3n)(a+2m)(2a-3n)(a+2m)(2a−3n)
分组分解法
定义
先分组,再分组分解,然后提公因式
应用
能分组分解的式子有四项或六项或大于六项,一般的分组分解有两种形式:二二分法,三一分法。
比如:
ax+ay+bx+byax+ay+bx+byax+ay+bx+by
=a(x+y)+b(x+y)a(x+y)+b(x+y)a(x+y)+b(x+y)
=(a+b)(x+y)(a+b)(x+y)(a+b)(x+y)
1.用分组分解法把 4x2−2x−y2−y4x^2-2x-y^2-y4x2−2x−y2−y 因式分解。
= (4x2−y2)−(2x+y)(4x^2-y^2)-(2x+y)(4x2−y2)−(2x+y)
= (2x−y)(2x+y)−(2x+y)(2x-y)(2x+y)-(2x+y)(2x−y)(2x+y)−(2x+y)
= (2x+y)(2x−y−1)(2x+y)(2x-y-1)(2x+y)(2x−y−1)
(1)在实数范围内分解因式: x5−64x^5-64x5−64 .
= x(x2+8)(x+22)(x−22)x(x^2+8)(x+2\sqrt 2)(x-2\sqrt 2)x(x2+8)(x+22)(x−22)
(2)已知 a,b,c,da,b,c,da,b,c,d为非负整数,且 ac+bd+ad+bc=1997ac+bd+ad+bc=1997ac+bd+ad+bc=1997 ,求 a+b+c+da+b+c+da+b+c+d 的值。
ac+bd+ad+bc=(a+b)(c+d)=1997ac+bd+ad+bc=(a+b)(c+d)=1997ac+bd+ad+bc=(a+b)(c+d)=1997
又因为1997为质数
不妨 a+b=1a+b=1a+b=1&&c+d=1997c+d=1997c+d=1997
a+b+c+d=1998a+b+c+d=1998a+b+c+d=1998
(3)已知 a,b,ca,b,ca,b,c 是一个三角形的三边,求证: a4+b4+c4−2a2b2−2b2c2−2c2a2<0a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2<0a4+b4+c4−2a2b2−2b2c2−2c2a2<0
= a4+b4+c4+2a2b2−2b2c2−2c2a2−4a2b2a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2-4a^2b^2a4+b4+c4+2a2b2−2b2c2−2c2a2−4a2b2
= (a2+b2−c2)2−4a2b2(a^2+b^2-c^2)^2-4a^2b^2(a2+b2−c2)2−4a2b2
= (a+b+c)(a+b−c)(a−b+c)(a−b−c)(a+b+c)(a+b-c)(a-b+c)(a-b-c)(a+b+c)(a+b−c)(a−b+c)(a−b−c)
由三角形两边之长大于第三边即判断正负
(4) 已知 x+y=3,x2+y2−xy=4x+y=3, x^2+y^2-xy=4x+y=3,x2+y2−xy=4 ,求 x4+y4+x3y+xy3x^4+y^4+x^3y+xy^3x4+y4+x3y+xy3 的值。
x4+y4+x3y+xy3x^4+y^4+x^3y+xy^3x4+y4+x3y+xy3
= x3(x+y)+y3(x+y)x^3(x+y)+y^3(x+y)x3(x+y)+y3(x+y)
= (x3+y3)(x+y)(x^3+y^3)(x+y)(x3+y3)(x+y)
= (x+y)2(x2−xy+y2)(x+y)^2(x^2-xy+y^2)(x+y)2(x2−xy+y2)
= 32⋅43^2\cdot432⋅4
= 12