八年级上册因式分解习题
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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
八年级上册因式分解习题
(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).
(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.分解因式:
(2)x10+x5-2;
(4)(x5+x4+x3+x2+x+1)2-x5.
2.分解因式:
(1)x3+3x2-4;
(2)x4-11x2y2+y2;
(3)x3+9x2+26x+24;
(4)x4-12x+323.
3.分解因式:
(1)(2x2-3x+1)2-22x2+33x-1;
(2)x4+7x3+14x2+7x+1;
(3)(x+y)3+2xy(1-x-y)-1;
(4)(x+3)(x2-1)(x+5)-20.