春七年级数学下册第一章整式乘除章末小结与提升课时作业北师大版

第一章整式的乘除章末小结与提高整式的运算种类1幂的运算23的结果是( )典例1计算(-3x)A.-3x5B.-27x6C.-3x6D.-27x5233236【分析】(-3x)=(-3)·(x)=-27x.【答案】B【针对训练】1.以下运算正确的选项是(A)A.x3·x2=x5B.(x3)3=x6C.x5+x5=x10D.x6-x3=x32.以下运算正确的选项是(C)A.-a4·a3=a7B.a4·a3=a12C.(a4)3=a12D.a4+a3=a71种类2整式的乘、除法运算典例2计算:5x2y÷(-xy)×(2xy2)2.【分析】原式=5x2y÷(-xy)×4x2y4=-15x×4x2y4=-60x3y4.【针对训练】1下边计算正确的选项是(C).A.3x2·4x=12x2B.x3·x5=x15C.x4÷x=x3D.(x5)2=x72.xy=7,则x-2y(A)若3=4,93的值为A.B.C.-3D.3.化简:6a6÷3a3=2a3.4.先化简,再求值:-2x3y4÷(-x2y2)·(-x)-(x-2y)(3y+x)+x(x+2xy2),此中x=-1,y=-2.解:原式=2xy2·(-x)-(x2-6y2+xy)+x2+2x2y2=-2x2y2-x2+6y2-xy+x2+2x2y2=6y2-xy.当x=-1,y=-2时,原式=6×(-2)2-(-1)×(-2)=22.种类3乘法公式1.若a2-b2=,a-b=,则a+b的值为(B)A.-B.C.1D.22.若(7x-a)2=49x2-bx+9,则|a+b|的值为45.3.先化简,再求值:[(x+2y)(x-2y)-(x+4y)2]÷4y,此中x2-8x+y2-y+16=0.解:x2-8x+y2-y+16=0,2即x2-8x+16+y2-y+=0,则(x-4)2+=0,则x-4=0且y-=0,解得x=4,y=.原式=[x2-4y2-(x2+8xy+16y2)]÷4y=(x2-4y2-x2-8xy-16y2)÷4y=-2x-5y.当x=4,y=时,原式=-8-=-.种类4整式的混淆运算典例3(邵阳中考)先化简,再求值:(a-2b)(a+2b)-(a-2b)2+8b2,此中a=-2,b=.【分析】原式=a2-4b2-a2+4ab-4b2+8b2=4ab,当a=-2,b=时,原式=-4.【针对训练】1.计算:(6x4-8x3)÷(-2x2)-(3x+2)(1-x).解:原式=-3x2+4x-3x+3x2-2+2x=3x-2.2.先化简,再求值:2(x+1)-(x+1)2,此中x=3.3解:原式=2x+2-(x2+2x+1)=2x+2-x2-2x-1=1-x2.当x=3时,原式=1-9=-8.3.先化简,再求值:[(a+b)(a-b)+(a+b)2-2a2]÷2a,此中a=3,b=-.解:原式=(a2-b2+a2+2ab+b2-2a2)÷2a=2ab÷2a=b.当a=3,b=-时,原式=-.种类5定义新运算41.在实数范围内定义运算“☆”,其规则为a☆b=a2-b2,则方程(4☆3)☆x=13的解为x=±6.2.现规定一种新的运算“※”:a※b=ba,如3※2=23=8,求4※和3※(-a2b).解:4※,3※(-a2b)=(-a2b)3=-a6b3.种类6规律研究1.依据以低等式:1=12,21+2+1=2,21+2+3+2+1=3,关于正整数n(n≥4),猜想1+2++(n-1)+n+(n-1)++2+1=n2.2.察看以下各式:3×5=15,15=42-1,5×7=35,35=62-1,511×13=143,