解下列方程(1)x(2x-7)=2x  ...
解下列方程(1)x(2x-7)=2x(2)x2-2x+4=0(3)(y+2)2=(3y-1)2(4)2y2+7y-3=0....
解下列方程(1)x(2x-7)=2x (2)x2-2x+4=0(3)(y+2)2=(3y-1)2 (4)2y2+7y-3=0.
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(1)x(2x-7)=2x
整理得:2x2-9x=0
x(2x-9)=0,
解得:x1=0,x2=4.5;
(2)x2-2x+4=0
配方得:(x-1)2=-3,
故此方程无实数根;
(3)(y+2)2=(3y-1)2
(y+2)2-(3y-1)2=0
[(y+2)+(3y-1)][(y+2)-(3y-1)]=0,
整理得:(4y+1)(-y+3)=0
解得:y1=-
,y2=3;
(4)2y2+7y-3=0
b2-4ac=49-4×2×(-3)=73,
故x=
,
则x1=
,x2=
.
整理得:2x2-9x=0
x(2x-9)=0,
解得:x1=0,x2=4.5;
(2)x2-2x+4=0
配方得:(x-1)2=-3,
故此方程无实数根;
(3)(y+2)2=(3y-1)2
(y+2)2-(3y-1)2=0
[(y+2)+(3y-1)][(y+2)-(3y-1)]=0,
整理得:(4y+1)(-y+3)=0
解得:y1=-
1 |
4 |
(4)2y2+7y-3=0
b2-4ac=49-4×2×(-3)=73,
故x=
?7±
| ||
4 |
则x1=
?7+
| ||
4 |
?7?
| ||
4 |
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