解:(1)x
2-2x=1,
x
2-2x+1=2,
(x-1)
2=2,
x-1=±
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,
∴x
1=1+
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,x
2=1-
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.
(2)x
2+2x=1,
x
2+2x+1=2,
(x+1)
2=2,
x+1=±
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,
∴x
1=-1+
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,x
2=-1-
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.
(3)△=1-4×(-1)=5,
x=
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,
∴x
1=
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,x
2=
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.
(4)△=9-4×(-1)=13,
x=
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,
∴x
1=
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,x
2=

.
(5)x
2+2x=1,
x
2+2x+1=2,
(x+1)
2=2,
x+1=±
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,
∴x
1=-1+
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,x
2=-1-
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.
(6)(x-3)
2=25,
x-3=±5,
∴x
1=8,x
2=-2.
(7)x
2-2
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x+2=0,
△=12-4×2=4,
x=
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=
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±1,
∴x
1=
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+1,x
2=
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-1.
(8)3(x
2-5)=4x,
3x
2-4x-15=0,
(3x+5)(x-3)=0,
∴x
1=-
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,x
2=3.
(9)3x
2+(x-2)=0,
3x
2+x-2=0
(3x-2)(x+1)=0,
∴x
1=
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,x
2=-1.
(10)(2x-1)(x+3)=4,
整理為2x
2+5x-7=0,
(2x+7)(x-1)=0,
∴x
1=-
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,x
2=-1.
(11)x
2-3x-4=0,
(x-4)(x+1)=0,
∴x
1=4,x
2=-1.
(12)x
2-3x-18=0,
(x+3)(x-6)=0,
∴x
1=-3,x
2=6.
分析:(1)利用配方法解方程;
(2)利用配方法解方程;
(3)利用公式法解方程;
(4)利用公式方法解方程;
(5)利用配方法解方程;
(6)先兩邊除以5,然后利用直接開平方法解方程;
(7)先把方程化為一般式,然后利用公式法解方程;
(8)先把方程化為一般式,然后利用因式分解法解方程;
(9)先把方程化為一般式,然后利用因式分解法解方程;
(10)先把方程化為一般式,然后利用因式分解法解方程;
(11)利用因式分解法解方程;
(12)利用因式分解法解方程.
點(diǎn)評:本題考查了解一元二次方程-因式分解法:先把方程右邊變形為0,再把方程左邊分解為兩個一次式的乘積,這樣原方程轉(zhuǎn)化為兩個一元一次方程,然后解一次方程即可得到一元二次方程的解.也考查了配方法、公式法解一元二次方程.