第1个回答 2010-07-26
1.用因式分解法解一元二次方程
(1)X²+3X+2=0
(x+1)(x+2)=0
x=-1或x=-2
(2)(2X+3)²-2(2X+3)=0
(2x+3)(2x+3-2)=0
(2x+3)(2x+1)=0
2x+3=0或2x+1=0
x=-3/2或x=-1/2
(3)(2Y+1)²-8(2Y+1)+15=0
(2y+1-3)(2y+1-5)=0
(2y-2)(2y-4)=0
2y-2=0或2y-4=0
y=1或y=2
2.用恰当的方法解下列方程
(1)(2X-3)²=9(2X+3)²
2x-3=3(2x+3)或2x-3=-3(2x+3)
2x-3=6x+9或2x-3=-6x-9
4x=12或8x=-6
x=3或x=-3/4
(2)X²-8X+6=0
公式法:
△=64-4*6=40
x=(8±√40)/2=4±√10
配方法:
x^2-8x+6=0
(x-4)^2=10
x-4=±√10
x=4±√10
这个题,怀疑你题目抄错了
如果是x^2-6x+8=0
那么(x-2)(x-4)=0
x=2或x=4
(3)(5X-1)²=3(5X-1)
(5x-1)(5x-1-3)=0
5x-1=0或5x-1-3=0
x=1/5或x=4/5
要详细过程,过程详细视情况加分本回答被提问者采纳
第2个回答 2010-07-26
(1)X2+3X+2=0
(X+1)(X+2)=0
X1=-1,X2=-2
(2)(2X+3)2-2(2X+3)=0
(2X+3)(2X+3-2)=0
(2X+3)(2X+1)=0
X1=-3/2,X2=-1/2
(3)(2Y+1)2-8(2Y+1)+15=0
(2Y+1-3)(2Y+1-5)=0
(2Y-4)(2Y-6)=0
Y1=2,Y2=3
(1)(2X-3)^2=9(2X+3)^2
(2X-3)^2-9(2X+3)^2=0
[(2X-3)+3(2X+3)][(2X-3)-3(2X+3)]=0
(8X+6)(-4X-12)=0
X1=-3/4,X2=-3
(2)X2-8X+6=0
(X-4)^2-16+6=0
(X-4)^2=10
X1=4+√10,x2=4-√10
(3)(5X-1)2=3(5X-1)
(5X-1)^2-3(5X-1)=0
(5X-1)(5X-1-3)=0
(5X-1)(5X-4)=0
X1=1/5,X2=4/5
第3个回答 2010-07-26
X²+3X+2=0
(X+2)(X+1)=0
X+2=0或者X+1=0
X1=-2 X2=-1
(2X+3)²-2(2X+3)=0
(2X+3)(2X+3-2)=0
(2X+3)(2X+1)=0
2X+3=0或者2X+1=0
X1=-3/2 X2=-1/2
(2Y+1)²-8(2Y+1)+15=0
(2Y+1-3)(2Y+1-5)=0
(2Y-2)(2Y-4)=0
2Y-2=0或者2Y-4=0
Y1=1 Y2=2
(2X-3)²=9(2X+3)²
(2X-3)²-9(2X+3)²=0
[2X-3+3(2X+3)][2X-3-3(2X+3)]=0
(2X-3+6X+9)(2X-3-6X-9)=0
(8X+6)(-4X-12)=0
8X+6=0或者-4X-12=0
X1=-3/4 X2=-3
X²-8X+6=0
X²-8X+16=10
(X-4)²=10
X-4=±√10
X1=4+√10 X2=4-√10
(5X-1)²=3(5X-1)
(5X-1)²-3(5X-1)=0
(5X-1)(5X-1-3)=0
5X-1=0或者5X-4=0
X1=1/5 X2=4/5