9.1～9.4 习题课-2021-2022学年七年级下册数学【拉网提优大小卷】苏科版

(2)原式=15b+3a2-10b2-2ab=3a2+13ab-1 (2)通过观察可以得到结论:≤2时,n1<(n+1)”; (3)原式=6x2+15xy-4xy-10y2=6x2+11xy-10y2 1>2时,n+1>(n+1) (4)原式=2(x2 (3)由(2)得到结论:201>2 ∴20122>2020 8.1~8.3习题课 12.(1)原式=3x2-3x.(2)原式=x2-14x-2.(3)原 式=22a-23.(4)原式=x2-x+2y 1.A2.C3.D4.A5.C6.a37.58.4 13.由(x+a)(x+2)=x2+(a+2x+2a,得a+2=-5,2a=b, 9.(a+b)5(2m-n)510. 解得a=-7,b=-14,则a+b=-21. 11.(1)原式 (2)原式=-p2q,(3)原式= 14.(1)由题意得(2x+4)(2x-5+4)=(2x+4)(2x-1) (4)原式=1+3+1-2=3. 12.(1)原式=-12+12=0.(2)原式=m3+m2-m3=m 答:活动场地扩大后的面积是(4x2+6x-4)m (3)原式=2a2-a14÷:a2=2a12-a2=a2.(4)原式 (2)4x2+6-4-2x(2x-5)=4x2+6x-4-4x2+10x p-q 13.(1)164解析:am”=am×a"=8×2=16;amn=am x=20时,原式=16×20-4=316 答:活动场地扩大后增加的面积是316m2 (2)∵am=8=23,a"=2,∴a"=(am)3=a3m,∴,m与n的 9.1~9.3习题课 数量关系为m=3 2.C3.B4.C5.D6.2x2y7 14.(1)1243=1015,468=101 8.-29.-9a2+3ab+2b210.M>N (2)(ab)c与a⊕(bc)不相等.理由如下: 11.(1)原式=4x3-6x2-2x+3.(2)原式=6x2+x-2 ∵(ab)c=10+c=1010×10=10 (3)原式=2x2-14xy+24y2.(4)原式=6m2 aG(c)=aF10+c=10-×101=10+10 11m+4n2.(5)原式=4x2-1.(6)原式=x3+y ∴(a⊕)⊕c与a⊕(bGc)不相等. 12.(1)原式=3×6b a'bc 第9章整式乘法与因式分解 (2)原式=4x4y2·(-3xy)=-12x5y3. 9.1~9.2习题课 (3)原式 xy2·(y2-2xy+x2)=~xy2+ 8.3.95×109.-x5y10.-20 11.(1)原式 =-8× (4)原式=-kx2y3·3xy2·4x2y2 b=a b2 (5)原式=9x4y2 (2)原式 b° )原式=2a3+60ab-8a (3)原式=9x4y 3nx2+m:x+3x2-9x+3m=x4+(n-3)x3+(m-3 (4)原式=-6x-2x2+8 3)x2+(mn-9)x+3m 12.(1)原式=x2-2x-x2+5x=3x 展开式中不含x2和x3项, (2)原式=-2x3y·3xy2 m-3+2=0.解得/=, -3=0, (3)原式=3a3+2a2-12 当m=6,n=3时,(n-m)=(3-6)3 (4)原式=( 14.(1)由题意得(3a+b)(2a+b)-(a+b)2=6a2+3ab 13.由题意得原多项式为x2 x+1+3x2=4x2 2ab+b-a2-2ab-b=5a2+3ab 答:绿化的面积是(5a2+3ab) 1,故正确计算结果应为一3x2·(4x2-2 (2)当a=2,b=4时,原式=20+24=4 答:绿化的面积是44m2 12x+2x2-3x2 9.4习题课(1) 14.原式=6a3-12a2+9a-6a3-8a2=-20a2+9a,当a 1.B2.C3.B4.B5. 2时,原式=-20×4-9×2=-98. 8.x3-x9.4a2+4ab+b210. 9.3习题课 1.(1)原式=9m2+30m2+25m 5.B6.2x2+x-37. (2)原式=(-b-a)(-b+a)=b2-a2. 8.19.210.x2+5x+6 (3)原式=a2+6a+9+a2-1-4a-8=2a2+2a. 11.(1)原式=a3+a2b+ab2-a2b-ab2-b2=a3-b3 (4)原式=9x2-4-(4x2-4x+1)=9x2-4-4x2 (3)原式=2(x2-1)-2x2+x=2x2-2-2x2+x=x-2. 12.(1)原式=(100-2)2=1002-2×100×2+22 (4)原式=4x2-4xy+y2+4(y2-x2)=5y2-4xy 10000-400+4=9604. 12.(1)原式=x2-2x-x+ (2)原式=152-(15+2)×(15-2)=152-152