用適當(dāng)方法解下列方程
(1)x2+4x-12=0
(2)x2-3x+2=0
(3)x(x-1)=x
(4)x2-3x+1=0
(5)(4x-1)2=(5x+2)2
(6)(2x+1)2+3(2x+1)+2=0.
【答案】
分析:(1)利用十字相乘法分解因式,即可得出答案;
(2)利用十字相乘法分解因式,即可得出答案;
(3)先移項,然后再提取公因式x即可得出答案;
(4)利用公式法直接求出方程的根即可;
(5)首先移項,再利用平方差公式分解因式,求出即可;
(6)將(2x+1)看做整體,利用十字相乘即可得出答案.
解答:解:(1)x
2+4x-12=0,
(x+6)(x-2)=0,
解得:x
1=2,x
2=-6;
(2)x
2-3x+2=0,
(x-1)(x-2)=0,
解得:x
1=2,x
2=1;
(3)x(x-1)=x,
x(x-1)-x=0,
x(x-1-1)=0,
解得:x
1=0,x
2=2;
(4)x
2-3x+1=0,
b
2-4ac=9-4=5>0,
x=
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=
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,
解得:x
1=
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,x
2=
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;
(5)(4x-1)
2=(5x+2)
2,
(4x-1)
2-(5x+2)
2=0,
[(4x-1)+(5x+2)][(4x-1)-(5x+2)]=0,
整理得出:(9x+1)(-x-3)=0,
解得:x
1=-
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,x
2=-3;
(6)(2x+1)
2+3(2x+1)+2=0,
[(2x+1)+2][(2x+1)+1]=0,
整理得出:(2x+3)(2x+2)=0,
解得:x
1=-
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,x
2=-1.
點評:本題主要考查了一元二次方程的計算方法,只有當(dāng)方程的一邊能夠分解成兩個一次因式,而另一邊是0的時候,才能應(yīng)用因式分解法解一元二次方程,難度適中.