PAGEPAGE1第一章整式的乘除章末小結(jié)與提升整式的運(yùn)算冪的運(yùn)算類型1冪的運(yùn)算典例1計算(-3x2)3的結(jié)果是()A.-3x5 B.-27x6C.-3x6 D.-27x5【解析】(-3x2)3=(-3)3·(x2)3=-27x6.【答案】B【針對訓(xùn)練】1.下列運(yùn)算正確的是(A)A.x3·x2=x5 B.(x3)3=x6C.x5+x5=x10 D.x6-x3=x32.下列運(yùn)算正確的是(C)A.-a4·a3=a7 B.a4·a3=a12C.(a4)3=a12 D.a4+a3=a7類型2整式的乘、除法運(yùn)算典例2計算:5x2y÷(-13xy)×(2xy2)2.【解析】原式=5x2y÷(-13xy)×4x2y=-15x×4x2y4=-60x3y4.【針對訓(xùn)練】1.下面計算正確的是(C)A.3x2·4x=12x2 B.x3·x5=x15C.x4÷x=x3 D.(x5)2=x72.若3x=4,9y=7,則3x-2y的值為(A)A.47 B.74 C.-3 D3.化簡:6a6÷3a3=2a3.

4.先化簡,再求值:-2x3y4÷(-x2y2)·(-x)-(x-2y)(3y+x)+x(x+2xy2),其中x=-1,y=-2.解:原式=2xy2·(-x)-(x2-6y2+xy)+x2+2x2y2=-2x2y2-x2+6y2-xy+x2+2x2y2=6y2-xy.當(dāng)x=-1,y=-2時,原式=6×(-2)2-(-1)×(-2)=22.類型3乘法公式1.若a2-b2=14,a-b=12,則a+b的值為(BA.-12 B.12 C.1 D2.若(7x-a)2=49x2-bx+9,則|a+b|的值為45.

3.先化簡,再求值:[(x+2y)(x-2y)-(x+4y)2]÷4y,其中x2-8x+y2-y+1614=0解:x2-8x+y2-y+1614=即x2-8x+16+y2-y+14=則(x-4)2+y-1則x-4=0且y-12=0,解得x=4,y=1原式=[x2-4y2-(x2+8xy+16y2)]÷4y=(x2-4y2-x2-8xy-16y2)÷4y=-2x-5y.當(dāng)x=4,y=12時,原式=-8-52=-類型4整式的混合運(yùn)算典例3(邵陽中考)先化簡,再求值:(a-2b)(a+2b)-(a-2b)2+8b2,其中a=-2,b=12.【解析】原式=a2-4b2-a2+4ab-4b2+8b2=4ab,當(dāng)a=-2,b=12時,原式=-4【針對訓(xùn)練】1.計算:(6x4-8x3)÷(-2x2)-(3x+2)(1-x).解:原式=-3x2+4x-3x+3x2-2+2x=3x-2.2.先化簡,再求值:2(x+1)-(x+1)2,其中x=3.解:原式=2x+2-(x2+2x+1)=2x+2-x2-2x-1=1-x2.當(dāng)x=3時,原式=1-9=-8.3.先化簡,再求值:[(a+b)(a-b)+(a+b)2-2a2]÷2a,其中a=3,b=-13解:原式=(a2-b2+a2+2ab+b2-2a2)÷2a=2ab÷2a=b.當(dāng)a=3,b=-13時,原式=-1類型5定義新運(yùn)算1.在實(shí)數(shù)范圍內(nèi)定義運(yùn)算“☆”,其規(guī)則為a☆b=a2-b2,則方程(4☆3)☆x=13的解為x=±6.

2.現(xiàn)規(guī)定一種新的運(yùn)算“※”:a※b=ba,如3※2=23=8,求4※13和3※(-a2b)解:4※133※(-a2b)=(-a2b)3=-a6b3.類型6規(guī)律探究1.依據(jù)以下等式:1=12,1+2+1=22,1+2+3+2+1=32,…對于正整數(shù)n(n≥4),猜想1+2+…+(n-1)+n+(n-1)+…+2+1=n2.

2.視察下列各式:3×5=15,15=42-1,5×7=35,35=62-1,…11×13=1