问题
解答题
计算下列各题 (1)3
(2)
(3)解方程:x2-3x+1=0. |
答案
(1)原式=(3
-22
)+(2
-33
)+(3
-33
)3
=
-22
.3
(2)原式=a2b• b a
=ab2
=ba
(3)∵b2-4ac=(-3)2-4=5
∴x=-b± b2-4ac 2a
=3± 5 2
即x1=
,x2=3+ 5 2
.3- 5 2
计算下列各题 (1)3
(2)
(3)解方程:x2-3x+1=0. |
(1)原式=(3
-22
)+(2
-33
)+(3
-33
)3
=
-22
.3
(2)原式=a2b• b a
=ab2
=ba
(3)∵b2-4ac=(-3)2-4=5
∴x=-b± b2-4ac 2a
=3± 5 2
即x1=
,x2=3+ 5 2
.3- 5 2