【答案】
分析:(1)移項后分解因式得出2x-3)(2x-3+1)=0,推出2x-3=0,2x-3+1=0,求出即可;
(2)分解因式得出(3x+1)(x-1)=0,推出3x+1=0,x-1=0,求出即可;
(3)根據(jù)平方差公式和完全平方公式展開,再合并即可;
(4)先化成最簡二次根式,再合并同類二次根式.
解答:解:(1)移項得:(2x-3)
2+(2x-3)=0,
(2x-3)(2x-3+1)=0,
2x-3=0,2x-3+1=0,
解得:x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/0.png)
,x
2=1;
(2)分解因式得:(3x+1)(x-1)=0,
3x+1=0,x-1=0,
解得:x
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/1.png)
,x
2=1;
(3)原式=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/2.png)
)
2-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/3.png)
)
2-(3-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/4.png)
+2)
=5-2-3+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/5.png)
-2
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/6.png)
-2;
(4)原式=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/7.png)
-5
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/8.png)
+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/9.png)
=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202804341630307/SYS201311032028043416303018_DA/10.png)
.
點評:本題考查了解一元二次方程和二次根式的混合運算,解(1)(2)小題的關(guān)鍵是能把一元二次方程轉(zhuǎn)化成一元一次方程,解(3)小題的關(guān)鍵是能靈活運用公式進(jìn)行計算.