问题
解答题
用适当的方法解下列方程(组): (1)(x-5)2-9=0; (2)3x2-1=6x (3)x2+2x-63=0 (4)
(5)
(6)
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答案
(1)(x-5)2-9=0;
(x-5)2=9,
∴x-5=±3,
∴x1=8,x2=2,
(2)3x2-1=6x,
∴3x2-6x-1=0,
△=b2-4ac=36+12=48,
x=
=6± 48 2×3 6±4 3 6
∴x1=1+
,x2=1-2 3 3
,2 3 3
(3)x2+2x-63=0,
∴(x-7)(x+9)=0,
∴x1=7,x2=-9,
(4)
-2x-1 x
=2,3x 2x-1
先设
=y,根据题意得:2x-1 x
y-
=2,3 y
∴y2-2y-3=0,
(y-3)(y+1)=0,
∴y-3=0或y+1=0,
∴y1=3,y2=-1,
∴
=3,或2x-1 x
=-1,2x-1 x
∴x1=-1,x2=
;1 3
把x1=-1和x2=
分别代入x-1中,都不等于0,1 3
∴x1=-1,x2=
是原方程的解;1 3
(5)
,x+y=13 ① (x-1)(y-1)=30 ②
由①得:x=13-y,
∴(13-y-1)(y-1)=30,
∴(12-y)(y-1)=30,
∴y2-13y+42=0,
(y-6)(y-7)=0,
∴y1=6,y2=7,
∴x1=13-6=7,x2=13-7=6,
∴
,x1=7 y1=6
;x2=6 y2=7
(6)
,x2-2xy-3y2=0 ① x-2y=6 ②
由②得:x=6+2y,
∴(6+2y)2-2(6+2y)y-3y2=0,
y2-4y-12=0,
(y-6)(y+2)=0,
∴y1=6,y2=-2,
∴x1=18,x2=2,
∴
,x1=18 y1=6
.x2=2 y2=-2