解下列方程:
(1)x2-2x+1=25
(2)1-8x+16x2=2-8x
(3)x2+5x+7=3x+11
(4)2x2+3x=3
【答案】
分析:(1)(2)運(yùn)用因式分解法解一元二次方程,應(yīng)先將方程的右邊化為0,左邊則化成兩個以此因式的乘積,令每個因式分別等于零,得到兩個一元一次方程,解這兩個一元一次方程,它們的解就是原方程的解;
(3)(4)運(yùn)用公式法求解,首先將方程的右邊化為0,再將各項的系數(shù)代入求根公式求解即可.
解答:解:(1)x
2-2x+1=25,
整理,得x
2-2x-24=0,
(x-6)(x+4)=0,
令x-6=0或x+4=0,
解,得x
1=6,x
2=-4.
(2)1-8x+16x
2=2-8x,
整理,得16x
2-1=0,
(4x-1)(4x+1)=0,
令4x-1=0或4x+1=0,
解,得x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/0.png)
,x
2=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/1.png)
.
(3)x
2+5x+7=3x+11,
整理,得x
2+2x-4=0,
解,得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/2.png)
=-1±
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/3.png)
即:x
1=-1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/4.png)
,x
2=-1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/5.png)
.
(4)2x
2+3x=3
整理,得2x
2+3x-3=0,
解,得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/7.png)
,
即:x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/8.png)
,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001154696010686/SYS201311030011546960106003_DA/9.png)
.
點評:本題主要考查運(yùn)用“因式分解法”和“公式法”解一元二次方程的能力,二者的相同點在于都應(yīng)首先將方程的右邊化為0.