(1)-3x2+22x-24=0
(2)(3x+5)2-4(3x+5)+3=0
(3)(3x+2)(x+3)=x+14
(4)6x2-x-12=0.
【答案】
分析:(1)利用公式法解方程;
(2)利用換元法解方程;
(3)、(4)利用因式分解法解方程.
解答:解:(1)∵方程-3x
2+22x-24=0的二次項(xiàng)系數(shù)a=-3,一次項(xiàng)系數(shù)b=22,常數(shù)項(xiàng)c=-24,
∴x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/2.png)
,
∴x
1=6,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/3.png)
;
(2)設(shè)3x+5=y,則原方程變?yōu)?br />y
2-4y+3=0,
∴(y-1)(y-3)=0,
解得,y=1或y=3;
①當(dāng)y=1時(shí),3x+5=1,解得x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/4.png)
;
②當(dāng)y=3時(shí),3x+5=3,解得,x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/5.png)
;
∴原方程的解是x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/6.png)
,或x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/7.png)
;
(3)由原方程,得
(x+4)(3x-2)=0,
解得x=-4,或x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/8.png)
;
(4)由原方程,得
(2x-3)(3x+4)=0,
解得,x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/9.png)
,或x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200545660042990/SYS201311032005456600429020_DA/10.png)
.
點(diǎn)評(píng):本題考查了解一元二次方程--公式法、因式分解法、換元法.應(yīng)根據(jù)方程的特點(diǎn)選擇解方程的方法.