问题
解答题
选用适当的方法解下列方程: (1)4(y-1)2=36 (2)x2-3x-2=0. (3)(2x+1)2-5(2x+1)+6=0 (4)3x2-6x-1=0 (配方法) (5)化简求值:(1+
|
答案
(1)4(y-1)2=36,
2(y-1)=±6,
y1=4,y2=-2.
(2)x2-3x-2=0,
b2-4ac=(-3)2-4×1×(-2)=17,
x=
,3± 17 2×1
x1=
,x2=3+ 17 2
;3- 17 2
(3)(2x+1)2-5(2x+1)+6=0,
(2x+1-3)(2x+1-2)=0,
2x+1-3=0,2x+1-2=0,
x1=1,x2=
.1 2
(4)3x2-6x-1=0,
x2-2x=
,1 3
x2-2x+1=
+1,1 3
(x-1)2=
,4 3
x-1=±
,2 3 3
x1=
,x2=3+2 3 3
.3-2 3 3
(5)(1+
)÷1 x x2-1 x
=
×x+1 x x (x+1)(x-1)
=
,1 x-1
当x=
+1时,原式=2
=1
+1-12
.2 2