第1个回答 2009-06-17
1.已知2x^2-3xy-20y^2=0,且y不等于0,则x/y的值为______
(2x+5y)(x-4y)=0
2x+5y=0,x-4y=0
2x=-5y,x=4y
所以x/y=-5/2或x/y=4
2.用因式分解法解下列方程:
(1)x(2x-1)=3(1-2x)
x(2x-1)+3(2x-1)=0
(x+3)(2x-1)=0
x=-3,x=1/2
(2) x^2-3x-10=0
(x-5)(x+2)=0
x=5,x=-2
(3) (x-3)(x+7)=-9
x^2+4x-21+9=0
x^2+4x-12=0
(x+6)(x-2)=0
x=-6,x=2
(4)(3y-2)^2=(2y-3)^2
(3y-2)^2-(2y-3)^2=0
(3y-2+2y-3)(3y-2-2y+3)=0
(5y-5)(y+1)=0
y=1,y=-1
(5)(2y+1)^2-8(2y+1)+15=0
(2y+1-3)(2y+1-5)=0
(2y-2)(2y-4)=0
4(y-1)(y-2)=0
y=1,y=2
(6)(x-1)^2=2x(1-x)
(x-1)^2+2x(x-1)=0
(x-1)(x-1+2x)=0
(x-1)(3x-1)=0
x=1,x=1/3本回答被提问者采纳
第2个回答 2009-06-17
1.已知2x^2-3xy-20y^2=0,且y不等于0,则x/y的值为______
同时除以y^2
2(x/y)^2-3x/y-20=0
(x/y=t
2t^2-3t-20=0
(2t+5)(t-4)=0
t1=-5/2,t2=4
所以x/y的值为-5/2或4
2.用因式分解法解下列方程:
(1)x(2x-1)=3(1-2x)
(x+3)(2x-1)=0
x1=-3,x2=1/2
(2) x^2-3x-10=0
(x-5)(x+2)=0
x1=5,x2=-2
(3) (x-3)(x+7)=-9
x^2+4x-12=0
(x+6)(x-2)=0
x1=-6,x2=2
(4)(3y-2)^2=(2y-3)^2
(3y-2)^2-(2y-3)^2=0
[(3y-2)-(2y-3)][(3y-2)+(2y-3)]=0
(y+1)(5y-5)=0
y1=-1,y2=1
(5)(2y+1)^2-8(2y+1)+15=0
2y+1=t
t^2-8y+15=0
(t-3)(t-5)=0
t1=3,t2=5
y1=1,y2=2
(6)(x-1)^2=2x(1-x)
(x-1)^2-2x(1-x)=0
(x-1)(x-1+2x)=0
(x-1)(3x-1)=0
x1=1,x2=1/3
两道题都写详细过程 谢谢了
第3个回答 2009-06-17
1. 2x^2-3xy-20y^2=(2x+5y)(x-4y)=0
2x=-5y x/y=-5/2
或者 x=4y x/y=4
所以答案=-5/2或4
2.(1) x(2x-1)=-3(2x-1)
x=-3
(2) x^2-3x-10=0
(x-5)(x+2)=0
x=5或x=-2
(3)(x-3)(x+7)=-9
x^2+4x-12=0
(x+6)(x-2)=0
x=-6或x=2
(4) )(3y-2)^2=(2y-3)^2
9y^2-12y+4=4y^2-12y+9
5y^2=5
y=1或y=-1
(5)(2y+1)^2-8(2y+1)+15=0
(2y+1-5)(2y+1-3)=0
(2y-4)(2y-2)=0
y=2或者y=1
(6) (x-1)^2=2x(1-x)
(x-1)(x-1+2x)=0
x=1或x=1/3
第4个回答 2009-06-17
1、2x^2-3xy-20y^2=(2x+5y)(x-4y)=0
x-4y=0--x/y=4
2x+5y=0--->x/y=-2.5
故而结果为4或者-2.5
2、
(1)x(2x-1)=3(1-2x)
(2x-1)(x-3)=0,x=3,x=0.5
(2) x^2-3x-10=0
(x-5)(x+2)=0,x=5,x=-2
(3) (x-3)(x+7)=-9
x^2+4x-21+9=0
x^2+4x-12=0
(x+6)(x-2)=0
x=-6,x=2
(4)(3y-2)^2=(2y-3)^2
9y^2-12y+4=4y^2-12y+9
5y^2=5
y^2=1
y=1,y=-1
(5)(2y+1)^2-8(2y+1)+15=0
(2y+1-3)(2y+1-5)=0
(y-1)(y-2)=0
y=1,y=2
(6)(x-1)^2=2x(1-x)
(x-1)(x-1+2x)=0
(x-1)(3x-1)=0
x=1,x=1/3