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  八年级上册因式分解习题

  八年级上册因式分解习题

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2020-03-30 18:40
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孙智勇

  (1)-2x5n-1yn+4x3n-1yn+2-2xn-1yn+4;

  (2)x3-8y3-z3-6xyz;

  (3)a2+b2+c2-2bc+2ca-2ab;

  (4)a7-a5b2+a2b5-b7.

  解(1)原式=-2xn-1yn(x4n-2x2ny2+y4)

  =-2xn-1yn[(x2n)2-2x2ny2+(y2)2]

  =-2xn-1yn(x2n-y2)2

  =-2xn-1yn(xn-y)2(xn+y)2.

  (2)原式=x3+(-2y)3+(-z)3-3x(-2y)(-Z)

  =(x-2y-z)(x2+4y2+z2+2xy+xz-2yz).

  (3)原式=(a2-2ab+b2)+(-2bc+2ca)+c2

  =(a-b)2+2c(a-b)+c2

  =(a-b+c)2.

  (4)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)

  分解因式:

  (1)x9+x6+x3-3;

  (2)(m2-1)(n2-1)+4mn;

  (3)(x+1)4+(x2-1)2+(x-1)4;

  (4)a3b-ab3+a2+b2+1.

  解(1)将-3拆成-1-1-1.

  原式=x9+x6+x3-1-1-1

  =(x9-1)+(x6-1)+(x3-1)

  =(x3-1)(x6+x3+1)+(x3-1)(x3+1)+(x3-1)

  =(x3-1)(x6+2x3+3)

  =(x-1)(x2+x+1)(x6+2x3+3).

  (2)将4mn拆成2mn+2mn.

  原式=(m2-1)(n2-1)+2mn+2mn

  =m2n2-m2-n2+1+2mn+2mn

  =(m2n2+2mn+1)-(m2-2mn+n2)

  =(mn+1)2-(m-n)2

  =(mn+m-n+1)(mn-m+n+1).

  (3)将(x2-1)2拆成2(x2-1)2-(x2-1)2.

  原式=(x+1)4+2(x2-1)2-(x2-1)2+(x-1)4

  =〔(x+1)4+2(x+1)2(x-1)2+(x-1)4]-(x2-1)2

  =〔(x+1)2+(x-1)2]2-(x2-1)2

  =(2x2+2)2-(x2-1)2=(3x2+1)(x2+3).

  (4)添加两项+ab-ab.

  原式=a3b-ab3+a2+b2+1+ab-ab

  =(a3b-ab3)+(a2-ab)+(ab+b2+1)

  =ab(a+b)(a-b)+a(a-b)+(ab+b2+1)

  =a(a-b)〔b(a+b)+1]+(ab+b2+1)

  =[a(a-b)+1](ab+b2+1)

  =(a2-ab+1)(b2+ab+1).

  (1)-2x5n-1yn+4x3n-1yn+2-2xn-1yn+4;

  (2)x3-8y3-z3-6xyz;

  (3)a2+b2+c2-2bc+2ca-2ab;

  (4)a7-a5b2+a2b5-b7.

  解(1)原式=-2xn-1yn(x4n-2x2ny2+y4)

  =-2xn-1yn[(x2n)2-2x2ny2+(y2)2]

  =-2xn-1yn(x2n-y2)2

  =-2xn-1yn(xn-y)2(xn+y)2.

  (2)原式=x3+(-2y)3+(-z)3-3x(-2y)(-Z)

  =(x-2y-z)(x2+4y2+z2+2xy+xz-2yz).

  (3)原式=(a2-2ab+b2)+(-2bc+2ca)+c2

  =(a-b)2+2c(a-b)+c2

  =(a-b+c)2.

  本小题可以稍加变形,直接使用公式(5),解法如下:

  原式=a2+(-b)2+c2+2(-b)c+2ca+2a(-b)

  =(a-b+c)2

  (4)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)

  (1)x9+x6+x3-3;

  (2)(m2-1)(n2-1)+4mn;

  (3)(x+1)4+(x2-1)2+(x-1)4;

  (4)a3b-ab3+a2+b2+1.

  解(1)将-3拆成-1-1-1.

  原式=x9+x6+x3-1-1-1

  =(x9-1)+(x6-1)+(x3-1)

  =(x3-1)(x6+x3+1)+(x3-1)(x3+1)+(x3-1)

  =(x3-1)(x6+2x3+3)

  =(x-1)(x2+x+1)(x6+2x3+3).

  (2)将4mn拆成2mn+2mn.

  原式=(m2-1)(n2-1)+2mn+2mn

  =m2n2-m2-n2+1+2mn+2mn

  =(m2n2+2mn+1)-(m2-2mn+n2)

  =(mn+1)2-(m-n)2

  =(mn+m-n+1)(mn-m+n+1).

  (3)将(x2-1)2拆成2(x2-1)2-(x2-1)2.

  原式=(x+1)4+2(x2-1)2-(x2-1)2+(x-1)4

  =〔(x+1)4+2(x+1)2(x-1)2+(x-1)4]-(x2-1)2

  =〔(x+1)2+(x-1)2]2-(x2-1)2

  =(2x2+2)2-(x2-1)2=(3x2+1)(x2+3).

2020-03-30 18:43:04

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