用適當?shù)姆椒ń夥匠蹋?br />(1)(3x+1)2=(2x-3)2
(2)(x+2)2-3(x+2)+2=0
(3)x2-4x-3=0
(4)9(2x+3)2=4(2x-5)2.
【答案】
分析:(1)開方得出方程3x+1=2x-3,3x+1=-(2x-3),求出方程的解即可;
(2)分解因式得到(x+2-1)(x+2-2)=0,推出x+2-1=0,x+2-2=0,求出方程的解即可;
(3)求出b
2-4ac的值,代入公式x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202341022901932/SYS201311032023410229019018_DA/0.png)
求出即可;
(4)開方后得出方程3(2x+3)=2(2x-5),3(2x+3)=-2(2x-5),求出方程的解即可.
解答:解:(1)開方得:3x+1=±(2x-3),
∴3x+1=2x-3,3x+1=-(2x-3),
解得:x
1=-4,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202341022901932/SYS201311032023410229019018_DA/1.png)
.
(2)分解因式得:(x+2-1)(x+2-2)=0,
∴x+2-1=0,x+2-2=0,
解得:x
1=-1,x
2=0.
(3)x
2-4x-3=0,
b
2-4ac=(-4)
2-4×1×(-3)=28,
∴x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202341022901932/SYS201311032023410229019018_DA/2.png)
=2±
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202341022901932/SYS201311032023410229019018_DA/3.png)
,
∴x
1=2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202341022901932/SYS201311032023410229019018_DA/4.png)
,x
2=2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202341022901932/SYS201311032023410229019018_DA/5.png)
.
(4)開方得:3(2x+3)=±2(2x-5),
∴3(2x+3)=2(2x-5),3(2x+3)=-2(2x-5),
解得:x
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202341022901932/SYS201311032023410229019018_DA/6.png)
,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202341022901932/SYS201311032023410229019018_DA/7.png)
.
點評:本題考查了解一元一次方程和解一元二次方程的應(yīng)用,根據(jù)是能把一元二次方程轉(zhuǎn)化成一元一次方程,主要培養(yǎng)了學生的計算能力和理解能力.