(2002•荆门)阅读下列范例,按要求解答问题.
例:已知实数a、b、c满足a+b+2c=1,a
2+b
2+6c+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/0.png)
=0,求a、b、c的值.
解法1:由已知得a+b=1-2c,①(a+b)
2-2ab+6c+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/1.png)
=0.②
将①代入②,整理得4c
2+2c-2ab+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/2.png)
=0.∴ab=2c
2+c+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/3.png)
③
由①、③可知,a、b是关于t的方程t
2-(1-2c)t+2c
2+c+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/4.png)
=0④的两个实数根.
∴△=(1-2c)
2-4(2c
2+c+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/5.png)
≥0,即(c+1)
2≤0.而(c+1)
2≥0,∴c+l=0,c=-1,
将c=-1代入④,得t
2-3t+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/6.png)
=0.∴t
1=t
2=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/7.png)
,即a=b=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/8.png)
.∴a=b,c=-1.
解法2∵a+b+2c=1,∴a+b=1-2c、设a=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/9.png)
+t,b=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/10.png)
-t.①
∵a
2+b
2+6c+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/11.png)
=0,∴(a+b)
2-2ab+6c+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/12.png)
=0.②
将①代入②,得(1-2c)
2-2
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/13.png)
+6c+
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/14.png)
=0.
整理,得t
2+(c
2+2c+1)=0,即t
2+(c+1)
2=0.∴t=0,c=-1.
将t、c的值同时代入①,得a=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/15.png)
,b=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/16.png)
.a=b=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/17.png)
,c=-1.
以上解法1是构造一元二次方程解决问题.若两实数x、y满足x+y=m,xy=n,则x、y是关于t的一元二次方程t
2-mt+n=0的两个实数根,然后利用判别式求解.
以上解法2是采用均值换元解决问题.若实数x、y满足x+y=m,则可设x=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/18.png)
+t,y=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131021232516037696942/SYS201310212325160376969004_ST/19.png)
-t.一些问题根据条件,若合理运用这种换元技巧,则能使问题顺利解决.
下面给出两个问题,解答其中任意一题:
(1)用另一种方法解答范例中的问题.
(2)选用范例中的一种方法解答下列问题:
已知实数a、b、c满足a+b+c=6,a
2+b
2+c
2=12,求证:a=b=c.
考点分析:
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