14章整式的乘法復習

1、第14章整式的乘法)()()().1 (3232aaaa1.計算：nn2)2(2).2(4)()()().3(235234aaaa5552:aaa原式解521422:nnn原式解46231012)(:aaaaa原式解232362)(2)( 3)().4(nnnnyxxyyx1852)125. 0().5(223322)5()3 . 0(5)6 . 0().6(baababbannnnnnnnyxyxyxyx62626262623:原式解82212)21(:18151853原式解5555552433243 . 95 . 78 . 1253 . 0536. 0:babababaababba原式解2

2、2232)(4)2(3).7(yxxxyxx)5)(1().8(2ttt)45)(32)(54)(32().9(xyyxyxyx325253256463:xyxyxxyx原式解5454)54(:2222ttttttt原式解4224222222524464)1625)(94(:yyxxxyyx原式解222)1 ()1)(1)(1 ().10(xxxx)64)(64().11(zyxzyx22222)3()3()3().12(xxx2222222222)11)(1 ()1 ()1)(1 (:xxxxxxx原式解22222364816)64()64()64(:zyzyxzyxzyxzyx原式解222

3、222222222222222236)33)(33()3()3()3)(3()3(:xxxxxxxxxx原式解)363025()65()964)(32().13(2222yxyxyxyxyxyx3333333333117243216125278)6()5()3()2(:xyyxyxyxyx原式解2. 己知10m=4 , 10n=5 , 求103m+2n 的值。160054)10()10(10101051041023232323nmnmnmnm解3. 先化簡，后求值:3x(-4x3y2)2-(2x2y)35xy 其中 x=1, y=2 .4. 己知x+5y=6 , 求 x2+5xy+30y 的值

4、。1282188404858163:474747473646yxyxyxxyyxyxx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)217215152612249124:22xxxxxxx解6173412205493)6(92012159:22xxxxxxxxx解.1,11:. 722的值求己知aaaa.1,51. 833的值求己知xxxx31,11121)1(11:22222aaaaaaaaaa故解110)11)(1(123125125122332222x

5、xxxxxxxxxxx故解 9. 己知 2x-3y=-4 , 求代數(shù)式4x2+24y-9y2 的值。10. 當x=-1 ,y=-2 時，求代數(shù)式 2x2-(x+y)(x-y)(-x-y)(-x+y)+2y2的值.16)4(4)32(424)32(424)32)(32(432yxyyxyyxyxyx原式解25) 2() 1()()(2)(22222222222222222yxyxyxyyxyxx原式解.1,31.1133的值求己知xxxx12. 計算：(x+1)(x+2)(x+3)(x+4)36)111 (3)11)(1(11119123122332222xxxxxxxxxxxx故解 24503

6、51024)5(10)5()65)(45(23422222xxxxxxxxxxxx原式解13. 計算：(a-1)(a4+1)(a2+1)(a+1)14. 計算：(2a-b)2(b+2a)2 11)() 1)(1() 1)(1)(1(82444422aaaaaaa原式解422422228164)2)(2(bbaabababa原式解15. 用科學記數(shù)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)(4222222222yxyxyxxyyxyxxyyxyxxyyxyxyxyx原式解17. 己

7、知x+y=3 ，x2+y2=5 則xy 的值等于多少？18. 己知x-y=4 , xy=21 ,則 x2+y2 的值等于多少？2459)(92929)(532222222xyyxxyyxyxyxyxyx故即解 582121621616216)(21422222xyyxyxyxyxxyyx即解 19. 根據(jù)己知條件，確定m ,n 的值 (a)己知：25m210n=5724 (b)己知: (x+1)(x2+mx+n) 的計算結(jié)果不含x2和x項。3241722525) 52 (251022547122nmnnmnnmnmnm解11001)() 1(23223nmnmmnxnmxmxnmxxnxmxx原式解 20. 己知：x+x-1=-3 , 求代數(shù)式x4+x-4 的值。 21計算：(a+b)(a+b)2-3ab(a-b)(a-b)2+3ab4749249)(79)(29)(34444222222112211xxxxxxxxxxxxxxxx故故故解66232333332222)()()()()()(:babababababababababa原式