数学：第13章整式的乘除复习课件（华东师大版八年级上）VIP免费

第13章整式的乘除)()()().1(3232aaaa1.计算：nn2)2(2).2(4)()()().3(235234aaaa5552:aaa原式解46231012)(:aaaaa原式解521422:nnn原式解232362)(2)(3)().4(nnnnyxxyyx1852)125.0().5(223322)5()3.0(5)6.0().6(baababbannnnnnnnyxyxyxyx62626262623:原式解82212)21(:18151853原式解5555552433243.95.78.1253.0536.0:babababaababba原式解22232)(4)2(3).7(yxxxyxx)5)(1().8(2ttt)45)(32)(54)(32().9(xyyxyxyx325253256463:xyxyxxyx原式解5454)54(:2222ttttttt原式解4224222222524464)1625)(94(:yyxxxyyx原式解222)1()1)(1)(1().10(xxxx)64)(64().11(zyxzyx22222)3()3()3().12(xxx2222222222)11)(1()1()1)(1(:xxxxxxx原式解22222364816)64()]64()][64([:zyzyxzyxzyxzyx原式解222222222222222222236)33)(33()3()3()]3)(3[()3(:xxxxxxxxxx原式解)363025()65()964)(32().13(2222yxyxyxyxyxyx3333333333117243216125278)6()5()3()2(:xyyxyxyxyx原式解2.己知10m=4,10n=5,求103m+2n的值。160054)10()10(10101051041023232323nmnmnmnm解3.先化简，后求值:3x(-4x3y2)2-(2x2y)3·5xy其中x=1,y=2.4.己知x+5y=6,求x2+5xy+30y的值。1282188404858163:474747473646yxyxyxxyyxyxx原式解36)5(630630)5(65:yxyxyyxxyx原式解6.解不等式：(3x+4)(3x-5)<9(x-2)(x+3)5.解方程：(2x-3)2=(x-3)(4x+2)217215152612249124:22xxxxxxx解6173412205493)6(92012159:22xxxxxxxxx解.1,11:.722的值求己知aaaa.1,51.833的值求己知xxxx31,11121)1(11:22222aaaaaaaaaa故解110)11)(1(123125125122332222xxxxxxxxxxxx故解9.己知2x-3y=-4,求代数式4x2+24y-9y2的值。10.当x=-1,y=-2时，求代数式[2x2-(x+y)(x-y)][(-x-y)(-x+y)+2y2]的值.16)4(4)32(424)32(424)32)(32(432yxyyxyyxyxyx原式解25])2()1[()())((]2)][(2[2222222222222222yxyxyxyyxyxx原式解.1,31.1133的值求己知xxxx12.计算：(x+1)(x+2)(x+3)(x+4)36)111(3)11)(1(11119123122332222xxxxxxxxxxxx故解2450351024)5(10)5()65)(45(23422222xxxxxxxxxxxx原式解13.计算：(a-1)(a4+1)(a2+1)(a+1)14.计算：(2a-b)2(b+2a)211)()1)(1()1)(1)(1(82444422aaaaaaa原式解42242222816]4[)]2)(2[(bbaabababa原式解15.用科学记数法表示：0.000000046116.己知x+y=4,求x3+12xy+y3的值。81061.4原式解64)(4)2(41244412)(412))((4222222222yxyxyxxyyxyxxyyxyxxyyxyxyxyx原式解17.己知x+y=3，x2+y2=5则xy的值等于多少？18.己知x-y=4,xy=21,则x2+y2的值等于多少？2459)(92929)(532222222xyyxxyyxyxyxyxyx故即解582121621616216)(21422222xyyxyxyxyxxyyx即解19.根据己知条件，确定m,n的值(a)己知：25m·2·10n=57·24(b)己知:(x+1)(x2+mx+n)的计算结果不含x2和x项。3241722525)52(251022547122nmnnmnnmnmnm解11001)()1(23223nmnmmnxnmxmxnmxxnxmxx原式解20.己知：x+x-1=-3,求代数式x4+x-4的值。21．计算：(a+b)[(a+b)2-3ab](a-b)[(a-b)2+3ab]4749249)(79)(29)(34444222222112211xxxxxxxxxxxxxxxx故故故解66232333332222)()())(())()()((:babababababababababa原式解433221])()()[(21)(:.22222222zyxxzzyyxzyx其中化简求值245813183213121)43()43()32()32(21)222222(21)(:222222xzyzxyzxyzxyzyxzyx原式解