【答案】
分析:(1)分類討論:當(dāng)k=0,方程變?yōu)椋簒-1=0,解得x=1;當(dāng)k≠0,△=(k+1)
2-4×k×(k-1)=-3k
2+6k+1,則-3k
2+6k+1≥0,利用二次函數(shù)的圖象解此不等式得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/0.png)
≤k≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/1.png)
;最后綜合得到當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/2.png)
≤k≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/3.png)
時(shí),方程有實(shí)數(shù)根;
(2)分類討論:當(dāng)k=0,方程變?yōu)椋簒-1=0,解得方程有整數(shù)根為x=1;當(dāng)k≠0,△=(k+1)
2-4×k×(k-1)=-3k
2+6k+1=-3(k-1)
2+4,要使一元二次方程都是整數(shù)根,則△必須為完全平方數(shù),得到k=1,2,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/4.png)
,k=1±
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/5.png)
;然后利用求根公式分別求解即可得到k=1、2、-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/6.png)
時(shí)方程的解都為整數(shù).
解答:解:(1)當(dāng)k=0,方程變?yōu)椋簒-1=0,解得x=1;
當(dāng)k≠0,△=(k+1)
2-4×k×(k-1)=-3k
2+6k+1,
當(dāng)△≥0,即-3k
2+6k+1≥0,方程有兩個(gè)實(shí)數(shù)根,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/7.png)
≤k≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/8.png)
,
∴當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/9.png)
≤k≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/10.png)
時(shí),方程有實(shí)數(shù)根;
(2)當(dāng)k=0,方程變?yōu)椋簒-1=0,解得方程有整數(shù)根為x=1;
當(dāng)k≠0,△=(k+1)
2-4×k×(k-1)=-3k
2+6k+1=-3(k-1)
2+4,
一元二次方程都是整數(shù)根,則△必須為完全平方數(shù),
∴當(dāng)△=4,則k=1;當(dāng)△=1,則k=2;當(dāng)△=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/11.png)
時(shí),k=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/12.png)
;當(dāng)△=0,則k=1±
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/13.png)
;
而x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/14.png)
,
當(dāng)k=1,解得x=0或-2;
當(dāng)k=2,解得x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/15.png)
或-1;
當(dāng)k=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/16.png)
,解得x=2或4;
當(dāng)k=1±
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/17.png)
,解得x都不為整數(shù),并且k為其它數(shù)△為完全平方數(shù)時(shí),解得x都不為整數(shù).
∴當(dāng)k為0、1、-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162914920532753/SYS201310221629149205327011_DA/18.png)
時(shí)方程都是整數(shù)根.
點(diǎn)評(píng):本題考查了一元二次方程ax
2+bx+c=0(a≠0)的根的判別式△=b
2-4ac:當(dāng)△>0,方程有兩個(gè)不相等的實(shí)數(shù)根;當(dāng)△=0,方程有兩個(gè)相等的實(shí)數(shù)根;當(dāng)△<0,方程沒有實(shí)數(shù)根.也考查了分類討論思想的運(yùn)用以及一元二次方程都為整數(shù)根的必要條件就是判別式為完全平方數(shù).