【答案】
分析:(1)設(shè)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/0.png)
=y,則原方程化為y+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/2.png)
,即 2y
2-5y+2=0,求出y
1=2,y
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/3.png)
,當(dāng)y
1=2時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/4.png)
=2,當(dāng)y
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/5.png)
時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/7.png)
,求出x即可;
(2)由②得出y=-2x,x=2y,將y=-2x代入①求出x
1=1,x
2=-1,y
1=-2,y
2=2,把x=2y代入①求出y
3=1,y
4=-1,x
3=2,x
4=-2,即可得出方程組的解.
解答:(1)解:設(shè)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/8.png)
=y,
則原方程化為y+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/10.png)
,
去分母得 2y
2-5y+2=0,
解得:y
1=2,y
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/11.png)
,
當(dāng)y
1=2時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/12.png)
=2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/13.png)
=4,
解得:x
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/14.png)
;
當(dāng)y
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/15.png)
時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/19.png)
,
解得:x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/20.png)
,
經(jīng)檢驗(yàn)x
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/21.png)
,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/22.png)
都是原方程的解.
(2)解:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/23.png)
由②得(2x+y)(x-2y)=0.
2x+y=0,x-2y=0,
y=-2x,x=2y,
將y=-2x代入①得:5x
2=5,
解得:x
1=1,x
2=-1,
即y
1=-2,y
2=2,
把x=2y代入①得:5y
2=5,
解得:y
3=1,y
4=-1,
即x
3=2,x
4=-2,
即原方程組的解為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/24.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/25.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/26.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192104257372199/SYS201311011921042573721027_DA/27.png)
.
點(diǎn)評:本題考查了解無理方程和解高次方程組,解無理方程的關(guān)鍵是能把無理方程轉(zhuǎn)化成有理方程,解高次方程組的關(guān)鍵是能把方程組轉(zhuǎn)化成解一元二次方程.