第二章 因式分解 知识点2 十字相乘法因式分解&知识点3 分组分解法因式分解-【荣恒专项】2026-2027学年八年级全一册数学——代数典型题专项训练

八年级代数题典型题专项训练 (6)原式=a2(x-y)-16(x-y) =(x-y)(a2-16) =(x-y)(a+4)(a-4); (7)原式=-2a(a2-6a+9) =-2a(a-3)2; (8)原式=(x2+1+2x)(x2+1-2x) =(x+1)2(x-1)2. 代数大冲关2 1.解：①+②得a2+3ab-2b2+b-3ab=a2 -b2=(a+b)(a-b). ①+③得a2+3ab-2b2+ab+6b2=a2+ 4ab+4b2=(a+2b)2. ②+③得b2-3ab+ab+6b2=7b2-2ab= b(7b-2a). 2.解：(1)令x+y=M,则(x+y)2-2(x+y)+ 1=M-2M+1=(M-1)2, .(x+y)2-2(x+y)+1=(x+y-1)2 故答案为(x+y-1)2; (2)令A=m+n,则(m+n)(m+n-4)+ 4=A(A-4)+4=A2-4A+4=(A-2)2, ∴.(m+n)(m+n-4)+4=(m+n-2)2; (3)(n+1)(n+2)(n2+3n)+1 =(n2+3n)[(n+1)(n+2)]+1 =(n2+3n)(n2+3n+2)+1 =(n2+3n)2+2(n2+3n)+1. 令n2+3n=A, 则原式=A2+2A+1 =(A+1)2 =(n2+3n+1)2 ,n是正整数，∴.n2+3n+1也为正整数， 102* ∴.式子(n+1)(n+2)(n2+3n)+1的值 一定是某一个整数的平方， 知识点2十字相乘法因式分解 代数大冲关 1.解：(1)x2+5x-14=(x+7)(x-2); (2)x2-2x-8=(x-4)(x+2); (3)x2-5x-24=(x+3)(x-8); (4)4x2-4x-15=(2x+3)(2x-5); (5)3x2-19x-14=(x-7)(3x+2); (6)(y2-y)2-14(y2-y）+24 =(y2-y-2)(y2-y-12) =(y-2)(y+1)(y-4)(y+3); （7)(x2+4x)2-2(x2+4x)-15 =(x2+4x-5)(x2+4x+3) =(x+5)(x-1)(x+3)(x+1); (8)x4-10x2+9 =(x2-1)(x2-9) =(x+1)(x-1)(x+3)(x-3); (9)2x2+5x-7=(2x+7)(x-1); (10)x2-2xy-8y2=(x+2y)(x-4y）; (11)-6a2-11a+7=-(6a2+11a-7) =-(2a-1)(3a+7); (12)32-12b-2762 =-(2762+12b-32) =-(3b+4)(9b-8). 知识点3分组分解法因式分解 代数大冲关 1.解：(1)x2-2xy+y2-25 =(x2-2xy+y2)-25 =(x-y)2-52 =(x-y+5)(x-y-5); (2)3x2-y-2y2-x+y =(3x2-xy-2y2)-(x-y) =(3x+2y)(x-y)-(x-y) =(x-y)(3x+2y-1); (3)2x3-2x2y+8y-8x =2x2(x-y）-8(x-y） =2(x-y）·x2-2(x-y）)×4 =2(x-y)(x2-4) =2(x-y)(x+2)(x-2); (4)x3+3x2y-4x-12 =(x3+3x2y）-(4x+12y) =x2(x+3y）-4(x+3y) =(x+3y)(x2-4)》 =(x+3y)(x+2)(x-2); (5)x2-2x-4y-4y2 =(x2-4y2)-(2x+4y) =(x+2y)(x-2y)-2(x+2y) =(x+2y)(x-2y-2); (6)5x2+6y-15x-2y =(5x2-15x)-(2y-6y） =5x(x-3)-2y(x-3) =(x-3)(5x-2y); (7)3x2-6y+3y2-27m =3(x2-2y+y2-9m2) =3[(x-y)2-(3m)2] =3(x-y+3m)(x-y-3m): (8)a2-2a+b2-2b+2ab+1 =(a2+2ab+b2)-(2a+2b)+1 =(a+b)2-2(a+b)+1 =(a+b-1)2; (9)通过观察本多项式，由2个4次单项 式，3个3次单项式，1个2次单项式组成， 参考答案与解析 把次数相近的分成一组. a"bc abed bc ab2 -ac2 c2d (a2bc abcd)+(bc-ab2)-(ac2 +c2d) =abe(a+d)+b(c-ab)-c2(a+d) =c(a+d)(ab-c)+b(c-ab) =(ab-c)[c(a+d)-b] =(ab-c)(ac+cd-b); (10)(a+b)(a+b-4)-c2+4 =(a+b)2-4(a+b)+4-c2 =(a+b-2)2-c2 =(a+b-2-c)(a+b-2+c). 知识点4因式分解的应用 代数大冲关1 1.【答案】D 【解析】a=2025x+2024,b=2025x+ 2025,c=2025x+2026, .∴.a-b=-1,b-c=-1,c-a=2, .'a2+b2+c2-ab-ac-bc =2(a2+b2+c2-ab-ac-bc）÷2 =(2a2+2b2+2c2-2ab-2ac-2bc)÷2 =[(a2-2ab+b2)+(b2-2bc+c2)+ (c2-2ac+a2)]÷2 =[(a-b)2+(b-c)2+(c-a)2]÷2 =[(-1)2+(-1)2+22]÷2 =6÷2 =3. 2.【答案】等边三角形 【解析】：a2+b2+c2-ab-bc-ac=0, .2a2+2b2+2c2-2ab-2bc-2ac=0, a2+b2-2ab+b2+c2-2bc+a2+c2-2ac=0. ∴.(a-b)2+(b-c)2+(c-a)2=0 4N103知识点2 ⊙知识点提炼： x2+(p+q)x+p9=(x+p)(x+q) 交叉相乘、十字验证 代数大冲关 1.分解因式： (1)x2+5x-14 (3)x2-5x-24 (5)3x2-19x-14 (7)(x2+4x)2-2(x2+4x)-15 (9)2x2+5x-7 (11)-6a2-11a+7 18* 上字相乘法因式分解 (2)x2-2x-8 (4)4x2-4x-15 (6)(y2-y)2-14(y-y)+24 (8)x4-10x2+9 (10)x2-2xy-8y2 (12)32-12b-27b2 第二章因式分解 知识点3分组分解法因式分解 ⊙知识点提炼： 方法点拨：分组分解法因式分解的原则： ①有公因式的分为一组；②按照系数配比分组； ③次数相近的分成一组；④按公式分组， 代数大冲关 1.因式分解： (1)x2-2xy+y2-25 (2)3x2-xy-2y2-x+y (3)2x3-2x2y+8y-8x (4)x3+3x2y-4x-12y (5)x2-2x-4y-4y2 (6)5x2+6y-15x-2xy (7)3x2-6xy+3y2-27m2 (8)a2-2a+b2-2b+2ab+1 (9)a'bc abed bc-ab2-ac2-c2d (10)(a+b)(a+b-4)-c2+4 N19