问题
解答题
(1)解方程:x2-2x-1=0; (2)计算:(3
|
答案
(1)移项,得:x2-2x=1,
配方:x2-2x+1=2,
即(x-1)2=2,
则x-1=±
,2
因而x1=
+1,x2=-2
+1;2
(2)原式=(3
-22
)[(33
-22
)-(33
+22
)]3
=(3
-22
)×(-43
)3
=-12
+24.6
(1)解方程:x2-2x-1=0; (2)计算:(3
|
(1)移项,得:x2-2x=1,
配方:x2-2x+1=2,
即(x-1)2=2,
则x-1=±
,2
因而x1=
+1,x2=-2
+1;2
(2)原式=(3
-22
)[(33
-22
)-(33
+22
)]3
=(3
-22
)×(-43
)3
=-12
+24.6