用適當的方法解方程:
①(2x-1)2=9;
②4x2-8x+1=0(配方法);
③2x-1)2=9(1-2x);
④3x2+5(2x+1)=1.
【答案】
分析:(1)利用直接開平方法的方法求解即可求得答案;
(2)利用配方法:先移項,再把二次項系數化為1,然后配方求解即可求得答案;
(3)利用因式分解法,提取公因式(2x-1),即可求得答案;
(4)首先整理,然后利用公式法求解即可求得答案.
解答:解:①∵(2x-1)
2=9,
∴2x-1=±3,
即2x-1=3或2x-1=-3,
解得:x
1=2,x
2=-1;
②∵4x
2-8x+1=0,
∴4x
2-8x=-1,
∴x
2-2x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/0.png)
,
∴x
2-2x+1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/1.png)
+1,
∴(x-1)
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/2.png)
,
解得:x
1=1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/3.png)
,x
2=1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/4.png)
;
③∵(2x-1)
2=9(1-2x),
∴(2x-1)
2+9(2x-1)=0,
∴(2x-1)(2x-1+9)=0,
即2x-1=0或2x-1+9=0,
解得:x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/5.png)
,x
2=-4;
④∵3x
2+5(2x+1)=1,
∴3x
2+10x+4=0,
∴a=3,b=10,c=4,
∴△=b
2-4ac=52,
∴x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/8.png)
,
解得:x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/9.png)
,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461636482/SYS201311032027504616364015_DA/10.png)
.
點評:此題考查了一元二次方程的解法.此題難度不大,解題的關鍵是選擇適當的解題方法.