七年级数学第七章整式的乘除习题讲解 试题VIP免费

第七章《整式的乘除》习题讲解 )()()().1(3232aaaa1. 计算：nn2)2(2).2(4)()()().3(235234aaaa5552:aaa原式解521422:nnn原式解46231012)(:aaaaa原式解 232362)(2)(3)().4(nnnnyxxyyx185 2)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3928)2()2(])2([)2().14(aaaa302)101()101()101().15(3333333333117243216125278)6()5()3()2(:xyyxyxyxyx原式解0)2()2(:3928aa原式解10099998)10(11001:3原式解 3333)31()31()3()3().16()83(1632).17(25454baabcba)4()7124().18(22323ababaa272)3()31()31()3(:3333原式解cbababca3825343)83(2:原式解2473:abba原式解 )](31[)](32)(2)[().19(23qpqpqpqp)37()949().20(22yxyx2. 己知 10m=4 , 10n=5 , 求 103m+2n 的值。2663632)(6)(3:222qpqpqpqpqp原式解yxyxyxyx3737)37)(37(:原式解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:...