初等数学专题研究答案(4)

（4）解方程： x?9x?10?x2?3x?10?0?x?5,x??2 x?69an?10?2an?1?2an?611(an?2)11an?2于是 ???an?1?59an?104(an?5)4an?5?5an?6an?211n?1a1?215?11n?22?4n311n?13?11n所以。所以an? ?n?1????nnn?1n3?11?11?4an?54a1?54411?4

第八讲 递归数列的通项（3）

求下列递推公式确定数列的通项公式：

?a1?1?a1?1 (2)? (1)??an?1?2an?3?an?1?2an?n??a1?1(3)? (4)na?2a?3?n?n?1??a1?3 ?na?a?2?n?n?1解：（1）设an?1?x?2(an?x)，则an?1?2an?x?x?3 所以an?3?(a1?3)?2n?1?4?2n?1?2n?1，an?2n?1?3

（2）设an?1?x(n?1)?y?2(an?xn?y)，则an?1?2an?xn?x?y 于是x?1,y?x?0?x?y?1，所以an?1?n?2?2(an?n?1) 所以an?n?1?(a1?1?1)?2n?1?3?2n?1，an?3?2n?1?n?1 （3）设an?1?x?3n?1?2(an?x?3n)，则an?1?2an?x?3n?x??1 所以an?3n?(a1?3)?2n?1??2?2n?1??2n，an?3n?2n

（4）设an?1?x(n?1)?2n?2(an?xn?2n?1)?an?1?2an?x?2n?x??1 所以an?n?2n?1?(a1?1)?2n?1?2?2n?1?2n，an?(n?2)2n?1

第九讲 高次方程的求根

解下列方程：

(1)6x4?13x3?12x2?13x?6?0

(2)30x4?17x3?228x2?17x?30?0 (3)15x5?34x4?15x3?15x2?34x?15?0 (4)3x4?7x3?7x?3?0 (5)2x4?9x3?9x?2?0

(6)x7?2x6?5x5?13x4?13x3?5x2?2x?1?0 (7)x6?5x4?5x2?1?0

(8)x6?3x5?2x4?6x3?2x2?3x?1?0 (9)x4?3x3?2x2?3x?1?0

解：（1）6x4?13x3?12x2?13x?6?0?6(x2?x?2)?13(x?x?1)?12?0

6(x?x?1)2?13(x?x?1)?0?x?x?1?0,x?x?1?32, 2313?x2?1?0,6x2?13x?6?0 6所以x??i,（2）30x4?17x3?228x2?17x?30?0?30(x2?x?2)?17(x?x?1)?228?0

82130(x?x?1)2?17(x?x?1)?168?0?x?x?1?,x?x?1??

3101253x2?8x?3?0,10x2?21x?10?0，所以x?3,?,,?

352（3）方程可以分解成(x?1)(x4?49x3?64x2?49x?15)?0 对于x4?49x3?64x2?49x?15?0，用x2去除两边：

15(x2?x?2)?49(x?x?1)?64?0?15(x?x?1)2?49(x?x?1)?34?0 [15(x?x?1)?34](x?x?1?1)?015(x?x?1)?34?0,x?x?1?1?0

15x2?34x?15?0,x2?x?1?0?(5x?3)(3x?5)?0,x???3i 2

53?1?3i?1?3i所以方程的根为：1,?,?, ,3522（4）3x4?7x3?7x?3?0?3(x2?x?2)?7(x?x?1)?0

3(x?x?1)2?7(x?x?1)?6?0?[3(x?x?1)?2](x?x?1?3)?0

3x2?2x?3?0,x2?3x?1?0?x?1?42i?3?5， ,x?32（5）2x4?9x3?9x?2?0?2(x2?x?2)?9(x?x?1)?0

?2(x?x?1)2?9(x?x?1)?4?0?[2(x?x?1)?1](x?x?1?4)?0

?2x2?x?2?0,x2?4x?1?0?x?1?17,x?2?2 4（6）原方程可以分解为(x?1)(x6?x5?6x4?7x3?6x2?x?1)?0

对于x6?x5?6x4?7x3?6x2?x?1得：(x3?x?3)?(x2?x?2)?6(x?x?1)?7?0

(x?x?1)(x2?x?2?1)?(x?x?1)2?2?6(x?x?1)?7?0 (x?x?1)[(x?x?1)2?3]?(x?x?1)2?2?6(x?x?1)?7?0

(x?x?1)3?(x?x?1)2?9(x?x?1)?9?0?(x?x?1?1)[(x?x?1)2?9]?0 x?x?1?1?0,x?x?1?3,x?x?1??3 解之得：x?3?5?1?3i?3?5；x?；x?

222?1?3i3?5?3?5；； 222所以原方程的根为: ?1;(7) 方程可以变形成(x2?1)(x4?4x2?1)?0

(3?1)2(6?2)2?对于x?4x?1?0得：x?2?3?x? 244222所以方程的根为：1,?1,6?2,26?2,22?66?2,? 22（8）方程可以分解成(x2?1)(x4?3x3?x2?3x?1)?0

对于x4?3x3?x2?3x?1?0?(x2?x?2)?3(x?x?1)?1?0

(x?x?1)2?3(x?x?1)?1?0?x?x?1??3?5 22x2?(3?5)x?2?0?x??3?5?30?65?3?5?30?65 ,x?44所以方程的根为1,?1,?3?5?30?65?3?5?30?65 ,44（9）方程可以变形成(x2?x?2)?3(x?x?1)?2?0?(x?x?1)2?3(x?x?1)?4?0

?x?x?1?3?7i?2x2?(3?7i)x?2?0 2而(3?7i)2?16?18?67i?3(7?27i?i2)?3(7?i)2?(21?3i)2 所以对于2x2?(3?7i)x?2?0得x?3?7i?(21?3i)，

4对于2x2?(3?7i)x?2?0得x?3?7i?(21?3i)

4所以方程的四个根为：3?21?(7?3)i3?21?(7?3)i，，

443?21?(7?3)i3?21?(7?3)i， 44第十讲 根式的化简与代数式值的计算

1、求下列各多项式的值：

(1)x?9?62,f(x)?2x3?20x2?23x?5 2(2)x?2?3?6,f(x)?x4?22x2?48x?2 (3)x?3?23?2,y?3?23?2,f(x,y)?3x2?5xy?3y2

解：（1）x?9?62?2x?9?62?4x2?36x?81?72?4x2?36x?9?0 2

所以f(x)?1137x(4x2?40x?46)?5?x(4x2?36x?9)?2x2??5 22237279?5??2x2?，而x2?9x? 224 ??2x2?927?18x?18?9(9?62)?18?99?542 所以f(x)??2(9x?)?42（2）x?2?3?6?(x?6)2?9?62?x2?3?26(x?3)

?x4?6x2?9?24x2?483x?72?x4?30x2?483x?63 所以f(x)?x4?22x2?48x?2?x4?30x2?8x2?48x?2

?8x2?483x?63?48x?2?8x2?48(3?1)x?65

?8(x2?3)?48(3?1)x?89?166(x?3)?48(3?1)x?89 ?(166?483?48)x?482?89

?1442?163?240?482?89?1922?163?329

（3）由x?3?23?2,y?3?23?2得xy?1,x?5?26,y?5?26,x?y?10，

所以f(x,y)?3x2?5xy?3y2?3(x?y)2?11xy?3?100?11?289 2、化简2?32?2?32?2?2?32?2?2?3 解：原式?2?32?2?34?(2?2?3 ?2?32?2?32?2?3?2?34?(2?3)

?2?32?3?4?3?1

3、化简下列各式

(1)4?12?37?40?5 (2)5?5?33?5?5?3 3解：（1）由于4?12?4?23?3?1，7?40?5?252?2?5?2

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